«      "♦ 


APPLIED  CALCULUS 

PRINCIPLES  AND  APPLICATIONS 

ESSENTIALS  FOR  STUDENTS 
AND    ENGINEERS 


BY 

ROBERT    GIBBES   THOMAS 

Professor  of  Mathematics  and  Engineering  at  the  Citadel, 
The  Military  College  of  South  Carolina 


45  EXERCISES    -    166  FIGURES 


NEW  YORK 

D.    VAN    NOSTRAND    COMPANY 

25  Park  Place 

1919 


Copyright,  1919, 

BY 

D.  VAN   NOSTRAND   COMPANY 


Stanbope  [Press 

H.  Gl  LSON    COM  PAN  Y 
BOSTON.  USA 


PREFACE 


This  book  as  a  first  course  in  the  Calculus  is  not  designed 
to  be  a  complete  exposition  of  the  Calculus  in  either  its 
principles  or  its  applications.  It  is  an  effort  to  make  clear 
the  basic  principles  and  to  show  that  fundamental  ideas  are 
involved  in  familiar  problems.  While  formulas  and  alge- 
braic methods  are  necessary  aids  to  concise  and  formal 
presentation,  they  are  not  essential  to  the  expression  of  the 
principles  and  underlying  ideas  of  the  Calculus.  These  can 
be  expressed  in  plain  language  without  the  use  of  symbols 
—  one  writer  challenging  the  citing  of  a  single  instance 
where  it  cannot  be  done. 

The  practice  is  common,  at  least  with  "  thoughtless  think- 
ers," of  blindly  using  formulas  without  an}r  true  conception 
of  the  ideas  for  which  they  are  but  the  symbolic  expres- 
sion. The  formulas  of  the  Calculus  are  an  invaluable  aid  in 
economy  of  thought,  but  their  effective  use  is  dependent 
upon  an  adequate  knowledge  of  their  derivation.  The 
object  of  this  book  is  to  set  forth  the  methods  of  the  Cal- 
culus in  sflch  a  way  as  to  lead  to  a  working  and  fruitful 
knowledge  of  its  elements,  to  exhibit  something  of  its  power, 
and  to  induce  its  use  as  an  efficient  tool.  No  claim  is  made 
for  absolute  rigor  in  all  the  deductions,  but  confidence  is 
invited  in  the  soundness  of  the  reasoning  employed  and  in 
the  logical  conclusions  obtained. 

There  are  students,  and  engineers  also,  who  when  con- 
strained to  use  the  Calculus  look  upon  it  as  a  necessary  evil. 
This  attitude  is  without  doubt  due  to  their  minds  having 
never  had  a  firm  grasp  upon  its  principles  nor  a  full  realiza- 

iii 

4 


iv  PREFACE 

tion  of  the  efficiency  of  its  methods.  A  student  while  tak- 
ing a  course  in  the  Calculus  usually  spends  at  least  one-half 
his  effort  in  reviewing  previous  mathematics.  In  fact,  a 
course  in  the  Calculus  is  held  to  involve  an  excellent  review 
of  geometry,  algebra,  and  especially  of  trigonometry;  hence, 
at  the  end  of  a  term  it  is  too  much  to  expect  of  the  average 
student  that  he  have  an  adequate  knowledge  of  the  Calculus. 

If  a  choice  must  be  made  between  the  ability  to  solve 
equations  (including  integration  processes)  and  the  far  more 
rare  ability  to  set  up  equations  to  represent  established  facts 
and  laws,  there  can  be  little  question  as  to  which  type  of 
ability  should  be  cultivated.  The  latter  is  of  higher  order 
and  is  likely  to  include  the  former.  Engineers,  physicists, 
inventors  and  men  of  science  generally  find  it  difficult  to 
translate  their  observations  into  language  which  the  pure 
mathematician  can  understand.  In  fact,  such  translation 
usually  involves  the  writing  of  the  equation:  an  undertaking 
beyond  the  capacity  equally  of  the  non-mathematical  scientist 
and  the  pure  mathematician.  Integration  of  the  equation, 
once  set  up,  the  mathematician  will  undertake;  conceiv- 
ably, so  might  a  machine.  Fruitful  deductions  and  rules 
of  practice  result.  The  difficulty  of  realizing  these  results 
arises  not  from  difficulties  in  moving  about  the  symbols,  but 
from  inability  on  the  part  of  nearly  all  persons  to  state  facts 
in  terms  of  symbols.  It  is  as  if  no  harmonist  knew  a  melody 
and  no  melodist  knew  a  note.  This  book  aims  to  keep  fact 
and  symbol  in  close  association,  so  that  the  student  will 
never  use  the  latter  without  being  conscious  of  the  former. 
It  may  then  be  expected  that  he  will  ultimately  be  able  to 
visualize  the  symbolic  expression  when  the  fact  is  known. 

Apart  from  the  references  in  the  text  and  in  footnotes, 
acknowledgment  is  here  made  of  the  clarifying  and  logical 
ideas  embodied  in  the  books  on  the  Calculus  by  Gibson,  by 
Taylor,  and  by  Tovmm md  and  Goodenough;  also  in  Hedrick's 
paper  on  the  Calculus  without  Symbols. 


PREFACE  V 

The  introduction  to  this  book  ends  with  a  reference  to 
the  discoverer  of  the  Calculus.  It  is  deemed  not  unfitting 
that  the  book  should  close  with  the  Central  Forces  of  the 
Principia. 

Robert  Gibbes  Thomas. 

The  Citadel,  Charleston,  S.  C. 
February  1st,  1919. 


CONTENTS 


INTRODUCTION. 
PART   I.     DIFFERENTIAL   CALCULUS. 


CHAPTER  I. 

FUNCTIONS.     DIFFERENTIALS.     RATES. 

Article  Page 

1 .  Variables  and  Constants 5 

2.  Functions.     Dependent  and  Independent  Variables 6 

3.  Function  —  Continuous  or  Discontinuous 7 

4.  Representation  of  Functional  Relation 7 

5.  Function  —  Increasing  or  Decreasing 8 

6.  Classes  of  Functions.     Empirical  Equations 9 

7.  Increments 12 

Exercise  1 13 

8.  Uniform  and  Non-uniform  Change 14 

9.  Differentials 15 

10.  Illustrations  of  Differentials 15 

11.  Rate,  Slope,  and  Velocity 18 

12.  Rate,  Speed,  and  Acceleration 19 

13.  Rate  and  Flexion 21 

14.  Illustrations 21 

Exercise  II 23 

CHAPTER   II. 
DIFFERENTIATION.     DERIVATIVES.     LIMITS. 

15.  Derivative 24 

16.  Differentiation 25 

17.  Limits 25 

18.  Theorems  of  Limits 26 

19.  Derivative  as  a  Limit.     Function  of  a  Function 26 

20.  Illustrative  Examples 31 

21.  Replacement  Theorem 37 

22.  Limit  of  Infinitesimal  Arc  and  Chord 37 

vii 


viii  CONTENTS 

ALGEBRAIC  FUNCTIONS. 

Article  Page 

23.   Formulas  and  Rules  for  Differentiation 38 

24r-31.   Derivation  of  Formulas 39-43 

Exercise  III 45 

LOGARITHMIC  AND  EXPONENTIAL  FUNCTIONS. 

32.  Formulas  and  Rules  for  Differentiation 47 

33.  Derivation  of  Formulas 48 

34.  Limit  ( 1  +  -X  =  e 50 

n=oo   \         nj 

35.  Derivation  of  Formulas 52 

36.  Limit  ( 1  +  -Y  =  ex 52 

37.  Derivation  of  Formulas 53 

38.  Modulus 54 

39.  Logarithmic  Differentiation 56 

40.  Relative  Rate.     Percentage  Rate 56 

Exercise  IV 57 

41.  Relative  Error 59 

42.  Compound  Interest  Law 60 

TRIGONOMETRIC  FUNCTIONS. 

43.  Circular  or  Radian  Measure 64 

44.  Formulas  and  Rules  for  Differentiation 64 

45.  Derivation  of  Formulas 65 

46.  Limit  /sin  d\       .,  a£f 

°±«  \~r)  =  1 66 

47-50.   Derivation  of  Formulas 68 

51.  Note  on  Formula 69 

52.  Remarks  on  Formula 70 

Exercise  V 71 

53.  The  Sine  Curve  or  Wave  Curve 72 

54.  Damped  Vibrations 73 

r 

INVERSE   TRIGONOMETRIC   FUNCTIONS. 

55.  Formulas  and  Rules  for  Differentiation 76 

56-59.    Derivation  of  Formulas 77,  78 

Exercise  VI 78 


CONTENTS  ix 

Article  Page 

60.  Hyperbolic  Functions 80 

61.  General  Relations 80 

62.  Numerical  Values.     Graphs 80 

63.  Derivatives 81 

64.  The  Catenary 81 

65.  Inverse  Functions 81 

66.  Derivatives  of  Inverse  Functions 82 

.      CHAPTER  III. 

SUCCESSIVE    DIFFERENTIATION.     ACCELERATION.     CURVI- 
*   LINEAR   MOTION. 

67.  Successive  Differentials 84 

68.  Successive  Derivatives 84 

69.  Resolution  of  Acceleration 86 

Exercise  VII 87 

70.  Circular  Motion 88 

71.  The  Second  Law  of  Motion 90 

72.  Angular  Velocity  and  Acceleration 92 

73.  Simple  Harmonic  Motion 94 

74.  Self -registering  Tide  Guage 97 

Exercise  VIII 97 


CHAPTER  IV. 
GEOMETRICAL  AND   MECHANICAL  APPLICATIONS. 

75.  (a)  Tangents  and  Normals 99 

(6)  Subtangents  and  Subnormals 100 

76.  Illustrative  Examples 101 

Exercise  IX 107 

77.  Polar  Subtangent,  Subnormal,  Tangent,  Normal 108 

Exercise  X 110 

CHAPTER  V. 
MAXIMA  AND   MINIMA.     INFLEXION  POINTS. 

78.  Maxima  and  Minima Ill 

79.  The  Condition  for  a  Maximum  or  a  Minimum  Value. . .  Ill 

80.  Graphical  Illustration 113 

81.  Rule  for  Applying  Fundamental  Test 115 


CONTENTS 

Article  Page 

82.  Rule  for  Determining  Maxima  and  Minima 115 

83.  Inclusive  Rule 116 

84.  Typical  Illustrations 118 

85.  Inflexion  Points 120 

86.  Polar  Curves 123 

87.  Auxiliary  Theorems 124 

Exercise  XI 125 

Problems  in  Maxima  and  Minima 127 

Determination  of  Points  of  Inflexion 132 

CHAPTER  VI. 
FRVATURE.     EVOLUTES. 


JUR^AI 


88.  Curvature V. 133 

89.  Curvature  of  a  Circle 134 

90.  Circle,  Radius,  and  Center  of  Curvature 135 

91.  Radius  of  Curvature  in  Rectangular  Coordinates 137 

92.  Approximate  Formula  for  Radius  of  Curvature 138 

Exercise  XII 139 

93.  Radius  of  Curvature  in  Polar  Coordinates 140 

Exercise  XIII 142 

94.  Coordinates  of  Center  of  Curvature 142 

95.  Evolutes  and  Involutes 143 

96.  Properties  of  the  Involute  and  Evolute 143 

97.  To  find  the  Equation  of  the  Evolute 144 

CHAPTER  VII. 

CHANGE    OF    THE   INDEPENDENT    VARIABLE.     FUNCTIONS 
OF  TWO   OR  MORE  VARIABLES. 

98.  Different  Forms  of  Successive  Derivatives 149 

99.  Change  of  the  Independent  Variable 149 

Exercise  XIV 151 

100.  Function  of  Several  Variables 152 

101.  Partial  Differentials 152 

102.  Partial  Derivatives 153 

103.  Tangent  Plane.     Angles  with  Coordinate  Planes 154 

Exercise  XV 156 

104.  Total  Differentials 156 

105.  Derivative  of  an  Implicit  Function 158 

Exercise  XVI 158 


CONTENTS  xi 

Abticle  Page 

106.  Total  Derivatives 159 

107.  Illustrative  Examples 160 

Exercise  X  VII 161 

108.  Approximate  Relative  Rates  and  Errors 162 

Exercise  XVIII 163 

109.  Partial  Differentials  and  Derivatives  of  Higher  Orders .  164 

110.  Interchange  of  Order  of  Differentiation 165 

Exercise  XIX 166 

111.  Exact  Differentials 167 

Exercise  XX 170 

112.  Exact  Differential  Equations 170 


PART  H.  INTEGRAL  CALCULUS 


CHAPTER  I. 
INTEGRATION.     STANDARD   FORMS. 

113.  Inverse  of  Differentiation 171 

114.  Indefinite  Integral 173 

115.  Illustrative  Examples 17-4 

116.  Elementary  Principles 178 

117.  Standard  Forms  and  Formulas , 180 

118.  Use  of  Standard  Formulas 182 

Exercise  XXI 185 

Exercise  XXII 187 

119-121.   Derivation  of  Formulas 187-190 

Exercise  XXIII 191 

122.  Reduction  Formulas 194 

123.  Integration  by  Parts 194 

Exercise  XXIV 197 

124.  Reduction  Formulas  for  Binomials 198 

Exercise  XXV 199 

CHAPTER  II. 
DEFINITE  INTEGRALS.     AREAS. 

125.  Geometric  Meaning  of  Jf(x)dx 204 

126.  Derivative  of  an  Area 205 

127.  The  Area  under  a  Curve   .    206 


Xll  CONTENTS 

Article                                                                                         •  Page 

128.  Definite  Integral.  . , 206 

129.  Positive  or  Negative  Areas 208 

130.  Finite  or  Infinite  Areas  —  "  Limits  "  Infinite 208 

131.  Interchange  of  Limits 210 

132.  Separation  into  Parts 210 

133.  Mean  Value  of  a  Function 211 

134.  Evaluation  of  Definite  Integrals 213 

Exercise  XXVI.. 213 

135.  Areas  of  Curves 214 

Exercise  XXVII 216 

136.  To  find  an  Integral  from  an  Area 219 

137.  Area  under  Equilateral  Hyperbola 221 

138.  Significance  of  Area  as  an  Integral 223 

139.  Areas  under  Derived  Curves 225 

CHAPTER  III. 

INTEGRAL   CURVES.    LENGTHS   OF   CURVES.    CURVE   OF  A 
FLEXIBLE   CORD. 

140.  Integral  Curves 227 

141.  Application  to  Beams 229 

142.  Lengths  of  Curves 236 

Exercise  XXVIII 237 

143.  Lengths  of  Polar  Curves 239 

Exercise  XXIX 240 

144.  Curve  of  a  Cord  under  Uniform  Horizontal  Load  — 

Parabola 241 

145.  The  Suspension  Bridge 244 

146.  Curve  of  a  Flexible  Cord  —  Catenary 245 

147.  Expansion  of  cosh  x/a  and  sinh  x/a 248 

148.  Approximate  Formulas 249 

149.  Solution  of  s  =  a  sinh  x/a 250 

150.  The  Tractrix 254 

151.  Evolute  of  the  Tractrix 256 

p~\         CHAPTER  IV. 

INTEGRATION  AS  THE  LIMIT  OF  A  SUM.     SURFACES  AND 

VOLUMES. 

152.  Limit  of  a  Sum 259 

153.  The  Summation  Process . 261 


CONTENTS  Xlll 

Abticle  Page 

154.  Approximate  and  Exact  Summations 262 

Exercise  XXX 266 

155.  Volumes 267 

156.  Representation  of  a  Volume  by  an  Area 268 

157.  Surface  and  Volume  of  any  Frustum 270 

Exercise  XXXI 282 

158.  Prismoid  Formula 283 

159.  Application  of  the  Prismoid  Formula 284 

Exercise  XXXII 287 

160.  Surfaces  and  Solids  of  Revolution 288 

Exercise  XXXIII 292 


CHAPTER  V. 

SUCCESSIVE   INTEGRATION.     MULTIPLE   INTEGRALS. 
SURFACES   AND   VOLUMES. 

161.  Successive  Integration X^^^^^^rT 296 

Exercise  XXXIV 299 

162.  Successive  Integration  with  Respect  to  Two  or  More 

Independent  Variables 300 

163.  The  Constant  of  Integration 301 

Exercise  XXXV 303 

164.  Plane  Areas  by  Double  Integration  —  Rectangular  Co- 

ordinates   304 

Exercise  XXXVI 307 

165.  Plane  Areas  by  Double  Integration  —  Polar  Coordi- 

nates   308 

Exercise  XXXVII 311 

166.  Area  of  any  Surface  by  Double  Integration 311 

Exercise  XXXVIII 316 

167.  Volumes  by  Triple  Integration  —  Rectangular  Coor- 

dinates    317 

Exercise  XXXIX 320 

168.  Solids  by  Revolution  by  Double  Integration 321 

169.  Volumes  by  Triple  Integration  —  Polar  Coordinates ..  .  321 

170.  Volumes  by  Double  Integration  —  Cylindrical  Coor- 

dinates    324 

171.  Mass.  Mean  Density 327 


xiv  CONTENTS 

CHAPTER  VI. 

MOMENT   OF  INERTIA.     CENTjER   OF   GRAVITY. 

Article  ^-^  \_^  Page 

172.  Moment  of  a  Force  about  an  Axis 331 

173.  First  Moments 331 

174.  Center  of  Gravity  of  a  Body 332 

175.  Center  of  Gravity  of  a  Plane  Surface 333 

176.  Center  of  Gravity  of  any  Surface 334 

177.  Center  of  Gravity  of  a  Line 334 

178.  Center  of  Gravity  of  a  System  of  Bodies 335 

179.  The  Theorems  of  Pappus  and  Guldin 336 

Exercise  XL 342 

180.  Second  Moments  —  Moment  of  Inertia 344 

181.  Radius  of  Gyration 345 

182.  Polar  Moment  of  Inertia 345 

183.  Moments  of  Inertia  about  Parallel  Axes 346 

184.  Product  of  Inertia  of  a  Plane  Area 348 

185.  Least  Moment  of  Inertia 349 

186.  Deduction  of  Formulas  for  Moment  of  Inertia 350 

187.  Moment  of  Inertia  of  Compound  Areas 352 

CHAPTER  VII. 
APPLICATIONS.     PRESSURE.     STRESS.     ATTRACTION. 

188.  Intensity  of  a  Distributed  Force 355 

189.  Pressure  of  Liquids 356 

Exercise  XLI 362 

190.  Attraction.     Law  of  Gravitation 363 

191.  Value  of  the  Constant  of  Gravitation 375 

192.  Value  of  the  Gravitation  Unit  of  Mass 376 

193.  Vertical  Motion  under  the  Attraction  of  the  Earth ....  376 

194.  Necessary  Limit  to  the  Height  of  the  Atmosphere 378 

195.  Motion  in  Resisting  Medium 379 

196.  Motion  of  a  Projectile 380 

197.  Motion  of  Projectile  in  Resisting  Medium 383 

CHAPTER  VIII. 
INFINITE   SERIES.     INTEGRATION   BY   SERIES. 

198.  Infinite  Series 385 

199.  Power  Series 386 


CONTENTS  XV 

Article  Page 

200.  Absolutely  Convergent  Series 388 

201.  Tests  for  Convergency 390 

Exercise  XLII 392 

202.  Convergency  of  Power  Series 393 

203.  Integration  and  Differentiation  of  Series 394 


CHAPTER  IX. 

TAYLOR'S   THEOREM.    EXPANSION   OF   FUNCTIONS. 
INDETERMINATE   FORMS. 

204.  Law  of  the  Mean 401 

205.  Other  Forms  of  the  Law  of  the  Mean 403 

206.  Extended  Law  of  the  Mean 405 

207.  Taylor's  Theorem 405 

208.  Another  Form  of  Taylor's  Theorem 406 

209.  Maclaurin's  Theorem 407 

210.  Expansion  of  Functions  in  Series 408 

211.  Another  Method  of  Deriving  Taylor's  and  Maclaurin's 

Series 408 

212-214.   Expansion  by  Maclaurin's  and  Taylor's  Theorems 411 

215.  Examples 412 

Exercise  XLII  I 418 

216.  The  Binomial  Theorem 419 

217.  Approximation  Formulas 421 

Exercise  XLIV 424 

218.  Application    of    Taylor's    Theorem    to    Maxima    and 

Minima 424 

219.  Indeterminate  Forms 425 

220.  Evaluation  of  Indeterminate  Forms 428 

221.  Method  of  the  Calculus 430 

Exercise  XLV 434 

222.  Evaluation  of  Derivatives  of  Implicit  Functions 435 


CHAPTER  X. 

DIFFERENTIAL  EQUATIONS.     APPLICATIONS.     CENTRAL 
w  FORCES. 

223.  Differential  Equations 436 

224.  Solution  of  Differential  Equations 436 

225.  Complete  Integral 437 


xvi  CONTENTS 

Article  Page 

226.  The  Need  and  Fruitfulness  of  the  Solution  of  Differ- 

ential Equations 441 

227.  Equations  of  the  Form  Mdx  +  Ndy  =  0 444 

228.  Variables  Separable 446 

229.  Equations  Homogeneous  in  x  and  y 446 

230.  Linear  Equations  of  the  First  Order 448 

231.  Equations  of  the  First  Order  and  nth  Degree 449 

232.  Equations  of  Orders  above  the  First 449 

233.  Linear  Equations  of  the  Second  Order 453 

APPLICATIONS. 

234.  Rectilinear  Motion 456 

235.  Curvilinear  Motion 460 

236.  Simple  Circular  Pendulum 461 

237.  Cycloidal  Pendulum 464 

238.  The  Centrifugal  Railway 466 

239.  Path  of  a  Liquid  Jet 467 

240.  Discharge  from  an  Orifice 469 

CENTRAL  FORCES. 

241.  Definitions 472 

242.  Force  Variable  and  Not  in  the  Direction  of  Motion . . .  472 

243.  Kepler's  Laws  of  Planetary  Motion 477 

244.  Nature  of  the  Force  which  Acts  upon  the  Planets 478 

245.  Newton's  Verification 480 

Index 483-490 


What  we  call  objective  reality  is,  in  the  last  analysis,  what  is  common 
to  many  thinking  beings  and  could  be  common  to  all;  this  common 
part  .  .  .  can  only  be  the  harmony  expressed  by  mathematical  laws. 

—  H.    POINCARE 


APPLIED    QALCULUS. 

s 

INTRODUCTION. 

The  Calculus  treats  of  the  rates  of  change  of  related 
variables.  The  factors  of  life  are  ever  changing,  acting  and 
reacting  upon  each  other.  The  quantities  with  which  we 
have  to  deal  in  ordinary  affairs  are  for  the  most  part  in  a 
state  of  change.  Hence  the  field  in  which  the  principles  of 
the  Calculus  are  directly  involved  is  a  wide  one. 

In  observing  the  changes  about  us  we  note  that  they  take 
place  at  various  rates,  and  the  determination  of  the  rapidity 
of  the  change  may  be  the  controlling  factor  in  many  investi- 
gations. Whenever  the  rapidity  of  the  change  of  anything 
is  in  question,  the  methods  of  the  Calculus  have  appropriate 
application. 

In  the  case  of  velocity  or  speed,  there  is  rate  of  change  of 
distance  and  time ;  in  a  thermometer  we  have  rate  of  change 
of  length  and  temperature,  while  in  the  barometer  there  is 
rate  of  change  of  height  and  density;  in  the  slope  of  ground 
or  grade  of  a  road  we  have  rate  of  change  of  vertical  height 
and  horizontal  distance;  and  in  the  case  of  a  curve,  rate  of 
change  of  ordinate  and  abscissa,  or  slope  of  the  curve. 

In  the  case  of  a  body  in  variable  motion,  it  becomes 
desirable  to  determine  its  velocity  at  some  point  of  its  path 
or  at  some  instant  of  time,  that  is,  the  instantaneous  velocity. 
This  notion  of  rate  of  change  at  an  instant  is  common  even 
to  untrained  minds. 

When  one  says  of  a  train  in  variable  motion,  that  it  is 
now  going  at  the  rate  of  sixty  miles  an  hour,  one  means  that 

1 


2  APPLIED  CALCULUS 

at  the  instant  considered  the  rate  is  such  that,  if  it  were 
maintained,  the  train  would  go  sixty  miles  in  an  hour,  that 
being  the  instantaneous  velocity. 

The  method  of  the  Calculus  in  getting  the  rate  of  change 
of  a  variable  at  any  instant  is  in  accordance  with  natural 
procedure:  measure  the  amount  of  change  in  a  short  period 
of  time,  then  the  average  rate  of  change  during  that  period 
is  the  ratio  of  amount  of  change  to  length  of  period;  the 
limit  approached  by  this  ratio,  as  the  period  of  time  is 
diminished  towards  zero  as  a  limit,  is  the  rate  of  change  at 
the  instant  the  period  began. 

In  determining  the  greatest  and  least  values  of  a  variable 
quantity,  they  are  found  where  the  rate  of  change  of  the 
variable  is  zero.  For  instance,  at  the  maximum  and  mini- 
mum temperature  during  a  day,  the  rate  of  change  of  the 
temperature  is  zero.  There  is  a  difference,  however,  in  that 
before  the  hottest  moment  the  temperature  was  rising  and 
then  afterwards  falling,  while  at  the  coldest  moment  the 
reverse  was  the  case.  In  both  cases  the  temperature's  rate 
of  change  was  momentarily  zero.  Here  is  to  be  seen  the 
method  of  the  Calculus  as  to  maxima  and  mimina. 

A  distinguishing  feature  of  the  Calculus  is  that  in  addition 
to  real  sensible  quantities  it  uses  ideal  hypothetical  concepts, 
which  are  quantities  that  exist  if  certain  conditions  are 
maintained.  The  Calculus  connects  these  two  classes  of 
quantities.  Passing  from  the  real  to  the  ideal  is  Differentia- 
tion, from  the  ideal  to  the  real  is  Integration.  The  advantage 
of  introducing  ideal  quantities  is  that  in  many  problems  an 
expression  for  the  ideal  is  readily  formed  and  from  this 
expression  the  real  quantity  is  obtained  by  Integration.  In 
other  cases  the  real  quantity  being  given,  the  problem  is 
solved  by  the  ideal  quantity,  obtained  from  the  real  by 
Differentiation. 

The  might  of  the  invisible  and  intangible  forces  in  Nature, 
predicated  upon  a  concept  (the  aether),  is  generally  recognized 


INTRODUCTION  3 

in  this  day  and  generation.  Therefore,  it  is  not  to  be 
wondered  at,  that  in  dealing  with  material  things  and  in 
seeking  the  inner  law  by  which  they  act  and  react  upon  each 
other,  we  should  call  to  our  aid  ideal  concepts.  The  exclu- 
sive realist  in  his  passion  for  facts  is  prone  to  overlook  the 
fact  that  ideas  are  the  first  of  facts. 

It  is  acknowledged  that  science  is  useless  unless  it  teaches 
us  something  about  reality.  Let  it  be  acknowledged  that 
the  aim  of  science  is  not  things  themselves,  but  the  relations 
between  things,  and  the  fruitfulness  of  the  ideal  quantities 
of  the  Calculus  is  recognized.  The  differentials  employed, 
when  properly  defined,  are  not  "  ghosts  of  departed  quan- 
tities," even  if  in  some  cases  ideal  in  character.  "Airy" 
perhaps,  but  never  "nothing"  they  give  to  the  creation  of 
the  mind  "a  local  habitation  and  a  name." 

While  the  ratio  of  some  real  quantities  may  never  equal 
the  ratio  of  the  ideal  quantities,  nevertheless  the  former 
ratio  may  approach  so  closely  the  latter  as  a  limit  that  the 
exact  value  of  the  ideal  ratio  can  be  discerned.  So  the 
differentiation  of  any  real  variable  quantity  is  possible.  On 
the  other  hand,  the  exact  integration  of  every  ideal  quantity 
is  not  possible,  for  in  some  cases  no  corresponding  real 
quantity  exists. 

In  this  respect  there  is  an  analogy  with  Involution  and 
Evolution.  Any  number  may  be  raised  to  a  power;  but 
the  exact  root  of  every  number  cannot  be  found,  for  no  real 
root  exists  for  some  numbers. 

Differential  Calculus  deals  with  the  rates  of  change  of 
continuous  variables  when  the  relation  of  the  variables  is 
known. 

Integral  Calculus  is  concerned  with  the  inverse  problem  of 
finding  the  relation  of  the  variables  themselves  when  their 
relative  rates  are  known. 

While  some  problems  to  which  the  Calculus  is  applied  may 
be  solved  by  other  methods,  it  often  furnishes  the  simplest 


4  APPLIED  CALCULUS 

solution;  and  there  are  cases  in  which  the  Calculus  alone 
gives  the  solution.  The  Calculus  is  a  tool  for  the  efficient 
worker,  and  in  the  hands  of  skillful  investigators  the  Cal- 
culus has  proved  to  be  a  powerful  instrument  in  bringing  to 
light  the  truths  of  Nature. 

In  reference  to  the  mighty  intellect  that  conceived  it,  there 
is  pardonable  hyperbole  in  the  lines  of  the  Poet :  — 

"  Nature  and  nature's  laws  lay  hid  from  sight, 
God  said,  'Let  Newton  be,'  and  all  was  light." 


ERRORS  AND  OMISSIONS 

Page  24,  on  last  line,  f(x)  =  m  for  f(x)  =  m. 
161,  in  Example  3,  35  for  36. 
174,  on  figure,  X  misplaced. 
196,  on  second  line,  integral  sign  omitted. 
212,  on  fifth  line,  CPX  for  CP. 
232,  on  fifth  line,  integral  sign  omitted. 
251,  Example*  1,  *Miller  and  Lilly's  Analytic  Mechanics. 
253,  at  bottom  of  page,  True  should  be  omitted. 
259,  on  second  line,  Art.  124  for  Art.  128. 
263,  at  end  of  fourth  line,  period  for  comma. 
345,  in  equation,  x~  for  y2. 
358,  in  (2),  x  for  x. 
385,  in  expansion,  factor  a  omitted. 

418,  in  Note,  -ir  for  x  in  (2)  gives  etir  =  1,  should  be,  -2  7r  for  x  in 
(2)    or  2tt  for  x  in  (1)  gives  e2lir  =  1,  whence  ei7r  =  ±1 
hence .  .  . 
456,  reference,  (Ex.  6,  Art.  116)  for  (Ex.  6,  Art.  115). 


PART  I. 

DIFFERENTIAL  CALCULUS. 


CHAPTER   I. 
FUNCTIONS.     DIFFERENTIALS.     RATES. 

1.  Variables  and  Constants.  —  A  variable  is  a  quantity 
that  changes  in  value.  It  is  said  to  vary  continuously  when, 
in  changing  from  one  value  to  another,  it  takes  each  inter- 
mediate value  successively  and  only  once.  If  at  any  value 
it  ceases  to  vary  continuously,  it  is  said  to  be  discontinuous 
at  that  value. 

A  constant  is  a  quantity  whose  value  is  fixed.  If  its  value 
is  always  the  same  in  every  discussion,  it  is  an  absolute 
constant.  If  the  fixed  value  may  be  different  in  different 
discussions,  it  is  an  arbitrary  constant. 

In  the  equation  of  the  circle,  x-  +  y2  =  r2,  x  and  y  are 
variables  that  vary  continuously  from  0  to  =t=r;  while  r  is  an 
arbitrary  constant,  as  its  value  is  fixed  only  for  any  one  circle. 

In  the  ordinary  affairs  of  life  we  have  to  deal  with  con- 
tinuous variables,  such  as  time;  the  distance  of  an  object  in 
continuous  motion  from  any  point  on  its  path;  and  with 
discontinuous  variables,  such  as  the  amount  of  a  sum  of 
money  at  interest  compounded  periodically;  the  price  of 
cotton;  the  cost  of  money  orders,  etc. 

In  nature  we  have  constants,  such  as :  thejnass  of  a  body, 
which  is  an  absolute  constant;  the  weight  of  a  body,  which 
is  an  arbitrary  constant,  as  it  is  fixed  according  to  latitude 

5 


6  DIFFERENTIAL  CALCULUS 

and  elevation;  in  mathematics,  the  ratio  of  the  circum- 
ference of  a  circle  to  its  diameter  and  the  base  of  Naperian 
logarithms  are  absolute  constants. 

Variables  are  represented  usually  by  the  last  letters  of  the 
alphabet;  as  x,  y,  z,  or  p,  6,  4>,  etc.  The  letter  A,  however, 
often  represents  a  variable  area. 

Absolute  constants  are  denoted  by  number  symbols,  and 
there  are  some  absolute  constants  represented  by  letters,  as 
7r,  e,  for  the  ratio  and  base  just  mentioned,  each  transcen- 
dental but  the  most  important  in  mathematics. 

Arbitrary  constants  are  represented  usually  by  the  first 
letters  of  the  alphabet;  as  a,  b,  c,  a,  0,  y,  etc.  Particular 
values  of  variables  are  constants  and  are  denoted  by  X\,  yi} 
zh  £2,  2/2,  Z2,  etc. 

2.  Functions.  Dependent  and  Independent  Vari- 
ables. —  When  two  variables  are  so  related  that  the  value 
of  one  of  them  depends  upon  the  value  of  the  other,  the  first 
is  the  dependent  variable  and  is  said  to  be  a  function  of  the 
second,  and  the  second  is  the  independent  variable,  which  in 
connection  with  the  function  is  usually  called  simply  the 
variable,  or  sometimes  the  argument. 

The  area  of  a  square  is  a  function  of  the  length  of  a  side. 
The  area  or  the  circumference  of  a  circle  is  a  function  of  its 
radius.  The  square,  or  the  square  root,  or  the  logarithm  of 
a  number,  is  a  function  of  the  number. 

Any  function  of  x  is  represented  by  /  (x),  F  (x),  $  (x),  etc., 

and  the  symbol  f  (x)  denotes  any  expression  involving  x, 

\    whose  value  depends  upon  the  value  of  x.     In  any  discussion 

involving  x,  f  (a)  means  the  value  of  /  (x)  when  x  is  replaced 

\xby  a  throughout  the  expression.     In  y  =  fix),  x  is  the 

\  J    independent  variable  and  y  is  the  function  or  dependent 

,  v  I    variable.     In  the  equation  x2  +  y2  =  r2,  y  =\^/r2  —  x2  or 

\f    -  x  =^\/r2  _  y2f  so  y  =  f(x)  or  x  =  f(y).     If  one  variable  is 

expressed  directly  in  terms  of  another,  the  first  is  said  to  be 

an  explicit  function  of  the  second.     If  the  relation  between 


REPRESENTATION  OF  FUNCTIONAL  RELATION  •     7 

the  two  variables  is  given  by  an  equation  containing  them 
but  not  solved  for  either,  then  either  variable  is  said  to  be 
an  implicit  function  of  the  other.  So  in  x2  +  y2  =  r2,  y  is 
an  implicit  function  of  x  and  x  is  an  implicit  function  of  y; 
but  in  y  =  Vr2  —  x2,  y  is  an  explicit  function  of  x,  and  in 
x  =  Vr2  —  y2,  x  is  an  explicit  function  of  y. 

A  variable  may  be  a  function  of  more  than  one  variable, 
thus  in  z2  —  x2  +  y2,  or  in  z  =  xy,  z  is. a  function  of  x  and  y. 
The  area  of  a  rectangle  is  a  function  of  its  base  and  altitude. 
The  volume  of  a  solid  is  a  function  of  its  three  dimensions; 
so  in  V  =  xijz,  V  =  f  (x,  y,  z). 

3.  Function  —  Continuous  or  Discontinuous.  —  A  func- 
tion as/  (x)  is  said  to  be  continuous  between  x  =  a  and  x  =  b, 
if  when  x  varies  continuously  from  a  to  b,  f  (x)  varies  con- 
tinuously from  f  (a)  to  /(&).  In  other  words,  f  (x)  is  con- 
tinuous between  x  =  a  and  x  =  b  when  the  locus  of  y  = 
f(x)  between  the  points  (a,  f  (a))  and  (b,  f  (6))  is  an  un- 
broken line,  straight  or  curved. 

A  function  is  said  to  be  discontinuous  at  any  value  when 
it  ceases  to  vary  continuously  at  that  value,  even  though  its 
variable  may  be  continuous. 

Some  functions  are  continuous  for  all  values  of  their 
variables;  others  are  continuous  only  between  certain 
limits.  For  example,  sin  0  and  cos  6  are  continuous  for 
values  of  6  from  6  t=  0  to  6  =  2  t;  tan  9  is  continuous  from 
,0  =  0  to  0  =  *72jind  from  6  =  tt/2  to  6  =  f  tt,  but  when 
^passes  througIT7r/2  or  |  r,  tan  6  changes  from  A-oo  to 
!—  go  ,  hence  tan  6  is  discontinuous  for  6  =  ir/2  or  |  t. 

The  Calculus  treats  of  continuous  variables  and  functions 
only,  or  of  variables  and  functions  between  their  limits  of 
continuity. 

4.  Representation  of  Functional  Relation.  —  Often  the 
relation  between  the  function  and  the  argument  can  be 
expressed  by  a  simple  formula.  For  example,  if  s  is  the 
distance  fallen  from  rest  in  time  t,  then  s  =  f  (t)  =  %  gt2. 


8  DIFFERENTIAL  CALCULUS 

In  such  cases,  the  value  of  the  function  for  any  value 
taken  for  the  variable  can  be  found  by  simply  substituting 
in  the  formula;   thus, 

si=/(l)  =ig,     s2=/(2)  =  ig.2*  =  2g, 
and  so  for  any  value. 

A  function  is  tabulated  when  values  of  the  argument,  as 
many  and  as  near  together  as  desired,  are  set  down  in  one 
column  and  the  corresponding  values  of  the  function  are  set 
down  opposite  in  another  column.  For  example,  in  a  table 
of  sines,  the  angle  in  degrees  and  minutes  is  the  argument, 
and  the  sine  of  the  angle  is  the  function. 

A  function  is  graphed  or  exhibited  graphically  by  laying 
off  the  values  of  the  argument  as  abscissas  along  a  horizontal 
axis,  and  at  the  end  of  each  abscissa  erecting  an  ordinate 
whose  length  will  represent  the  corresponding  value  of  the 
function;  a  curve  drawn  through  the  tops  (or  bottoms)  of 
the  ordinates  is  called  the  curve,  or  the  graph  of  the  function. 

If  V  =  f(x),  the  curve  is  the  locus  of  the  equation;  but 
it  is  the  length  of  the  ordinate  up  (or  down)  to  the  curve, 
rather  than  the  curve  itself,  that  represents  the  function. 

If  p  =  /(0),  the  function  may  be  graphed  by  laying  off  at 
a  point  on  a  line  as  axis  the  various  angles,  —  values  of  the 
argument  0,  and  along  the  terminal  sides  of  the  angles  the 
corresponding  values  of  the  function  p;  a  curve  through  the 
ends  of  the  vectorial  radii  will  be  the  graph  of  the  function, 
and  will  be  a  polar  curve.  Here,  too,  it  is  the  length  of  the 
radius  to  the  curve,  rather  than  the  curve  itself,  that  repre- 
sents the  function.  The  area  under  a  curve  may  be  taken 
to  represent  a  function  while  the  ordinate  or  radius  repre- 
sents some  other  function.     (See  Art.  139.) 

5.  Function  —  Increasing  or  Decreasing.  —  An  increas- 
ing function  is  one  that  increases  when  its  variable  increases, 
hence,  it  decreases  when  its  variable  decreases.  A  decreasing 
function  is  one  that  decreases  when  its  variable  increases, 
hence  it  increases  when  its  variable  decreases.     Thus  a!c  and 


CLASSES  OF   FUNXTIOXS  9 

ax  are  increasing,  and  ax  and  a  —  x  are  decreasing  functions 
of  x. 

6.  Classes  of  Functions.  —  An  algebraic  function  is  one 
that  without  the  use  of  infinite  series  can  be  expressed  by  the 
operations  of  addition,  subtraction,  multiplication,  division 
and  the  operations  denoted  by  constant  exponents. 

The  common  forms  are:  (a  ±  bx),  (a  =b  bxn),  ax,  a/x,  xn, 
including  x2,  x3,  \//x,  1/Vx,  etc. 

All  functions  which  are  not  algebraic  are  called  trans- 
cendental.    Of  these,  the  most  important  are: 

The  exponential  functions,  y  =  ax  or  bx,  and  y  —  e*,  and 
their  inverse  forms,  the  logarithmic  functions, 
x  =  loga  y   or   logi  y   and   x  =  loge  y. 

The  trigonometric  functions,  y  =  sin  6,  x  =  cos  6,  y  =  tan  0, 
and  the  inverse  trigonometric  functions,  d  =  arc  sin  y  or  sin-1  y, 
6  =  arc  cos  x  or  cos-1  x,  d  =  arc  tan  y  or  tan-1  y. 

The  hyperbolic  functions,  sinh  x  =  (ex  —  e~x)/2,  cosh  x  = 
(ex  +  e-1) /2,  tanh  x  =  (ex  —  e~x)  ex  +  er*);  and  the  inverse 
hyperbolic  functions, 

sinh-1  x  =  y  =  log  (x  +  Vx2  +  l), 
cosh-1  x  =  y  =  log  (x  d=  Vx2  —  l), 
tanh"1  x  =  y  =  \  log  (1  +  x/1  —  x). 

Note.  —  The  phenomena  of  change  in  Nature,  in  general, 
have  been  found  to  be  in  accordance  with  one  or  the  other 
of  three  fundamental  laws.  These  have  been  stated  *  to  be 
the  parabolic  law,  expressed  by  the  power  function  y  =  axn, 
where  n  is  constant,  positive  or  negative;  the  harmonic  law, 
expressed  by  the  periodic  function  y  =  a  sin  (mx) ;  and  the 
law  of  organic  growth,  or  the  compound  interest  law,  expressed 
by  the  exponential  function  y  =  aebx.  It  is  to  be  noted  that 
as  x  increases  in  arithmetic  progression,  y  of  the  exponential 
function  increases  in  geometrical  progression;    while,  as  x 

*  In  Elementary  Mathematical  Analysis,  by  Charles  S.  Slichter. 


10  DIFFERENTIAL  CALCULUS 

increases  in  geometrical  progression,  y  of  the  power  function 
increases  in  geometrical  progression  also. 

Examples  in  Nature  of  the  working  of  these  three  laws 
will  be  given  later. 

EMPIRICAL  EQUATIONS. 

Very  often  the  form  of  a  function  is  given  only  empirically; 
that  is,  the  values  of  the  function  for  certain  values  of  the 
variable  are  known  from  experiment  or  observation,  and  the 
intermediate  values  are  not  given;  for  example,  the  height 
of  the  tide  read  from  a  gauge  every  hour. 

In  such  cases  the  Calculus  is  not  of  much  use  unless  some 
known  mathematical  law  can  be  found  which  represents  the 
function  sufficiently  accurately. 

This  problem  of  finding  a  mathematical  function  whose 
graph  shall  pass  through  a  series  of  empirically  given  points  is 
of  great  practical  importance. 

The  known  values  of  the  function  and  of  the  variable  are 
plotted  on  cross-section  papeiW logarithmic  squared  paper" 
greatly  facilitating  the  solution,  and  a  smooth  curve  being 
drawn  "to  fit"  the  determined  points,  the  equation  of  this 
curve  is  required.  The  curve  suggested  by  the  plotted 
points  may  have  for  its  equation  one  of  the  following  forms: 

(straight  line)  y  =  a  +  bx,  or  y  =  ?nx; 

(parabola)  y  =  a  +  bx  +  ex2,  or  y  =  a  +  ex2; 

(hyperbola)  y  =  a  +  c/x  +  b,  y  =  l/xn,  or 

xy  =  bx  +  ay; 
(sine  curve)  y  =  a  sin  (bx  +  c) ,  or  y  =  a  sin  (mx) ; 

(power  function)  y  =  axn  (n  any  number) ; 

(exponential  function)  y  =  aebx. 

If  the  curve  suggested  by  the  plotted  points  is  a  straight 
line,  determine  the  values  of  a  and  b,  or  of  m,  from  the 
observed  data.  The  straight  line  is  not  likely  to  pass 
through  all  the  points  plotted,  even  when  the  straight-line 


EMPIRICAL  EQUATIONS  11 

law  is  the  correct  expression  of  the  relation  to  be  determined; 
for  the  experimental  data  are  subject  to  error.  If  the  line 
fits  the  points  within  the  limits  of  accuracy  of  the  experiment, 
it  may  be  drawn  through  two  of  the  plotted  points,  and  a 
and  b,  or  m,  may  be  evaluated  from  their  coordinates. 

By  appropriate  treatment  of  the  data  many  of  the  laws 
can  be  transformed  into  a  linear  relation.  Thus,  when  the 
points  plotted  suggest  a  vertical  parabola  with  its  vertex  on 
the  y-Sixis,  the  required  equation  will  be  of  the  form, 

y  =  a  +  ex2.  (1) 

If  t  is  put  for  x2  in  (1),  and  the  values  of  t  and  y  plotted, 
these  values  satisfy  the  relation  y  =  a  +  ct,  that  is,  a  straight- 
line  law.     The  power  function  y  =  axn  may  be  expressed : 

log  y  =  log  a -\- n  log  x,  (2) 

that  is,  the  logarithms  of  the  given  data  satisfy  a  straight-line 
law.  The  straight-line  law  to  fit  the  logarithms  can  be 
determined  and  compared  with  (2)  to  find  a  and  n,  which 
are  substituted  in  y  =  axn. 

The  hyperbolic  law  and  the  exponential  function  also  can 
be  transformed  to  the  straight-line  law,  and  the  constants 
evaluated.  Whether  the  experimental  data  can  be  expressed 
by  a  power  function  or  by  an  exponential  function  can  be 
determined  by  a  test.  When  the  data  show  that,  as  the 
argument  changes  by  a  constant  factor,  the  function  also 
changes  by  a  constant  factor,  then,  the  relation  can  be 
expressed  by  a  power  function. 

When,  however,  it  is  f  omuijhat-a-^hange  of  the  argument 
by  a  constant  increment  changes  the  function  by  a  constant 
factor,  then  the  relation  can  be  expressed  by  an  equation  of 
the  exponential  type.     (See  Note,  Art.  6.) 

A  full  discussion  of  this  problem  of  finding  the  expression 
of  the  relation  between  a  function  and  its  argument  from 
limited  experimental  data  involves  the  theory  of  least 
squares,  and  is  out  of  place  in  a  first  course  in  the  Calculus. 


12  DIFFERENTIAL  CALCULUS 

This  necessarily  inadequate  treatment  of  the  subject  here  is 
warranted  by  the  importance  of  the  problem. 

*  Example.  —  The  amount  of  water  Q,  in  cu.  ft.  that  flows 
through  100  feet  of  pipe  of  diameter  d,  in  inches,  with  initial 
pressure  of  50  lbs.  per  sq.  in.  is  given  by  the  following: 

d        1  1.5  2^3  4  6 

Q      4.88       13.43      27.50      75.13       152.51      409.54 

Find  a  relation  between  Q  and  d. 

Let  x  =  log  d,  y  =  log  Q ;  then  the  values  of  x  and  y  are : 

x  =  \ogd      0.000     0.176     0.301     0.477     0.602     0.788 
y  =  \ogQ      0.688     1.128     1.439     1.876     2.183     2.612 

These  values  plotted  give  points  in  the  xy  plane  very  nearly 
on  a  straight  line;  therefore,  taking  y  =  a  +  bx,  a  and  b  can 
be  evaluated  by  measurement  on  the  figure; 

a  =  0.688  =  log  4.88,     b  =  2.473. 

Hence,    log Q  =  log 4.88  +  2.473 log d  =  log  (4.88 d2m) ; 

whence  Q  =  4.88  d2-m.  *  (Ziwet  and  Hopkins.) 

7.  Increments.  —  The  amount  of  change  in  the  value  of 
a  variable  is  called  an  increment.  If  the  variable  is  increas- 
ing, its  increment  is  positive;  if  it  is  decreasing,  its  increment 
is  negative  and  is  really  a  decrement. 

An  increment  of  a  variable  is  denoted  by  putting  the  letter 
A  before  it;  thus  Az,  Ay  and  A  (a:2)  denote  the  increments 
of  x,  y,  and  x2,  respectively.  If  y  =  f(x),  Ax  and  ly  denote 
corresponding  increments  of  x  and  y,  and 

ky  =  Af(x)=f(x  +  Ax)-f(x), 

...      Ay  =  A/(s)=;/(s  +  As)-/(s) 
Ax         Ax  Ax 

x  denoting  any  value  of  x. 

In  the  figure,  let  OPi  ...  B  be  the  locus  of  y  =  f  (x) 
referred  to  the  rectangular  axes  OX  and  OY.     If  when  x  = 


INCREMENTS 


13 


OAfi,    Ax  =  M1M2,    then    At/  =  M2P2  -  M^  -  DP2;     if 
when  x  =0M3,  Ax  =  M3Mi}  then  Aj^MiP^MsPb  =  -#P8. 


^       A. 


In  the  last  case  Ay  is  negative  and  is  what  algebraically 
added  to  M3P3  gives  M4P4.     When 

x  =  QMX  =  Xi,  f  (x)  =  M 1P1  =  /  (zi)  ; 

when 

a;  =  0M2  =  Xl  +  Ax,    f  (x)  =  M2P2  =f(xi  +  Ax)  ; 
hence  when 

x  =  xh     A/  (x)  =  M2P2  -  MiPi  =  /  (a*  +  Ax)  -  /  (xO. 


EXERCISE  I. 

1.  One  side  of  a  rectangle  is  10  feet.     Express  the  variable  area  A 
as  a  function  of  the  other  side  x. 

2.  Express  the  circumference  of  a  circle  as  a  function  of  its  radius 
r;  of  its  diameter  d. 

3.  Express  the  area  of  a  circle  as  a  function  of  its  radius  r;   of  its 
diameter  d. 

4.  Express  the  diagonal  d  of  a  square  as  a  function  of  a  side  x. 

5.  The  base  of  a  triangle  is  10  feet.     Express  the  variable  area  A 
as  a  function  of  the  altitude  y. 

6.  If  y  =  f(x),   y+Ay  =  f(x+Ax); 

.-.     Ay  =Af(x)  =f(x+&x)-f(x), 
and  hence, 

Ay  =Af(x)  =  f(x+Ax)  -  f  (x) 
Ax          Ax  Ax 

If  y  =  mx  +  6,  find  value  of  Ay  and  of  -—' 

AX 


14  DIFFERENTIAL  CALCULUS 

Aw 

7.  If  y  =»  x2,  find  value  of  Ay  and  of  —  • 

8.  If  y  =  x3,  find  value  of  A?/  and  of  -£• 

9.  If  y  =  zn,  find  value  of  A?/  and  of  -^,  assuming  the  binomial 
theorem. 

10.   If  y  =  f  (x)  =  mx  +  6,  write  values  of 


/«)),  /a),  /(-i),  /(-^} 


8.  Uniform  and  Non-uniform  Change.  —  When  the  ratio 
of  the  corresponding  increments  of  two  variables  is  constant, 
either  variable  is  said  to  change  uniformly  with  respect  to 
the  other. 

When  y  =  mx  +  b,  -^-  =  m  (constant).     (6,  Exercise  I.) 

It  follows  that  any  linear  function  of  x  changes  uni- 
formly with  respect  to  x;  that  is,  y  changes  uniformly  with 
respect  to  x  when  the  point  (x,  y)  moves  along  any  straight 
line. 

When  the  ratio  of  the  corresponding  increments  of  two 

variables  is  variable,  either  variable  is  said  to  change  non- 

uniformly  with  respect  to  the  other. 

Aw 
When  y  =  x2,  -r^  =  2  x  +  Ax  (variable).     (7,  Exercise  I.) 

L\X 

Thus  the  area  of  a  square  changes  non-uniformly  with 
respect  to  a  side.  Any  non-linear  function  of  x  changes 
non-uniformly  with  respect  to  x,  for  evidently  y  changes  non- 
uniformly  with  respect  to  x  when  the  point  (x,  y)  moves 
along  any  curved  line. 

Since  time  changes  uniformly,  any  variable  will  change 
uniformly  when  it  receives  equal  increments  in  equal  times; 
and  it  will  change  non-uniformly  when  it  receives  unequal 
increments  in  equal  times. 

Thus  in  s  =  vt,  where  s  is  the  space  passed  over  in  time  t 


ILLUSTRATIONS  OF   DIFFERENTIALS 


15 


by  an  object  moving  with  constant  velocity  v,  s  changes 
uniformly. 

In  s  =  \  gt2,  where  the  object  moves  with  constant  accel- 
eration g,  s  changes  non-uniformly. 

9.  Differentials.  —  The  differentials  of  variables  that 
change  uniformly  with  respect  to  each  other  are  their  corre- 
sponding increments;  that  is,  their  actual  changes. 

The  differentials  of  variables  that  change  non-uniformly 
with  respect  to  each  other  are  what  would  be  their  corre- 
sponding increments  if,  at  the  corresponding  values  con- 
sidered, the  change  of  each  became  and  continued  uniform. 

As  with  increments,  the  differentials  will  be  positive  or 
negative  according  as  the  variables  are  increasing  or  de- 
creasing. 

The  differential  of  a  variable  is  denoted  by  putting  the 
letter  d  before  it;  thus,  dx,  read  " differential  x"  is  the 
symbol  for  the  differential  of  x.  The  differential  of  a  vari- 
able or  function  consisting  of  more  than  a  single  letter  is 
indicated  by  the  letter  d  before  a  parenthesis  enclosing  the 
variable  or  function;  thus,  d(x2),  d(mx-\-b),  d(f(x)), 
denote  the  differentials  of  x2,  mx  +  b,  and/(V),  respectively. 

10.  Illustrations  of  Differentials.  —  (a)  Suppose  a  rec- 
tangle, with  constant  altitude,  is  changing  by  the  base  in- 
creasing. If  when  the  base  is 
A  B  its  increase  is  BM,  then 
d  (base)  =  BM,  and  d  (rectangle) 
=  BMNC. 

Here  the  variables  change  uni- 
formly with  respect  to  each  other, 
hence  their  differentials  are  their 
corresponding  increments. 

(b)  Conceive  a  right  triangle,  with  variable  base  and 
altitude,  is  changing  by  the  altitude  moving  uniformly  to 
the  right.  If  when  the  base  is  A B  its  increment  is  BM,  then 
the  increment  of  the  triangle  will  be  BMDC.     But  if  the 


N 


16 


DIFFERENTIAL  CALCULUS 


increase  of  the  triangle  became  uniform  at  the  value  A  BC, 
the  increment  of  the  triangle  in  the  same  time  would  evi- 
dently be  BMNC;  hence,  BMNC  and  BM  may  be  taken  as 
the  differentials  of  the  triangle  and 
of  the  base,  where  the  base  is  A  B. 

In  this  case  the  triangle  changes 
non-uniformly  with  respect  to  its 
base,  so  its  differential  is  what  would 
be  its  increment  if,  at  the  value  con- 
sidered, the  change  became  uniform. 
Since  the  base  changes  uniformly,  its 
differential  is  its  actual  increment. 
Here  increment  of  triangle  ABC  = 
d  (triangle  ABC)  +  triangle  CND, 
while  A  (base)  =  d  (base).  If  the 
change  of  a  variable  be  uniform,  any  actual  increment  may 
be  taken  as  its  differential.  If  time  be  considered,  the  in- 
terval of  time,  though  arbitrary,  must  be  the  same  for  a 
function  as  for  its  variable. 

(c)  Let  the  curve  OPn  be  the  locus  of  y  =  f(x),  referred 
to  the  axes  OX  and  OY.  Conceive  the  area  between  OX 
and  the  curve  as  traced  by 
the  ordinate  of  the  curve 
moving  uniformly  to  the 
right.  Let  z  denote  this 
area,  and  let  MM\  be  Ax 
reckoned  from  the  value 
OM  =  x;  then  MM1P1P 
=  A3.  But  if  the  increase 
of  z  became  uniform  at 
the  value  OMP,  its  incre- 
ment in  the  same  interval 

would  be  MM^DP;  hence  MMX  and  MMXDP  may  be 
taken  as  the  differentials  of  x  and  z  respectively,  when 
x  =  OM. 


ILLUSTRATIONS  OF   DIFFERENTIALS 


17 


Hence  dz  =  MMXDP  =  MPdx  =  ydx, 

which  shows  that  area  z  is  changing  y  times  as  fast  as  x. 
Here  Iz  =  dz  +  area  PDPX. 

It  is  seen  here  that  while  the  actual  change  in  the  area 
does  not  admit  of  an  exact  geometrical  expression,  the 
differential  of  the  area,  being  a  rectangle,  is  exactly  and 
simply  expressed.  It  will  be  shown  further  on  how  by 
Integration  an  exact  expression  for  the  area  itself  is  obtained 
from  this  expression  for  the  differential. 

Note.  —  Historically  the  Calculus  originated  through  the 
efforts  to  obtain  the  exact  area  of  figures  boimded  by  curves, 
mathematics  up  to  that  time  having  furnished  no  method 
applicable  to  all  curves  whose  equations  were  known. 

It  is  true  too  that  historically  the  method  of  Integration 
was  discovered  before  the  method  of  Differentiation  was 
developed.  The  Differential  Calculus  arose  through  the 
problem  of  determining  the  direction  of  the  tangent  at  any 
point  of  a  curve.     (See  Note,  Art.  75.) 

(d)  Let  OPn  be  the  locus  of  y  =  /  0)  and  s  the  length 
from  0  along  the  curve.  Suppose  the  point  (x,  y)  to  move 
along  the  curve  to  P  and  thence 
along  the  tangent  at  that  point. 
Then  at  the  value  x  =  OM,  the 
change  of  x  and  y  would  become 
uniform  with  respect  to  each 
other,  as  the  point  (x,  y)  would 
be  moving  along  a  straight  line. 
The  change  of  s  would  become 
uniform  also  with  respect  to  both 
x  and  y.  As  x  is  the  independent 
variable  it  may  be  taken  to  vary 
uniformly,  making  PD  or  dx  = 
Az  or  MMi,  the  actorchange  in  x  as  the  point  moves  along 
the  curve  from  P  to  Pi.     Then  dy  is  DT,  the  corresponding 


18  DIFFERENTIAL  CALCULUS 

uniform  change  of  y,  and  ds  is  PT,  the  corresponding  uniform 
change  of  s.  It  is  evident  that  while  dx  =  Ax,  dy  is  not 
equal  to  Ay  and  ds  is  not  equal  to  As.  When,  and  only  when, 
the  locus  is  a  straight  line  will  dy  =  Ay  and  ds  =  As,  after  dx 
has  been  taken  equal  to  Ax. 

It  should  be  noted  that  it  is  not  essential  that  dx  should  be 
made  equal  to  Ax,  for  dx  may  be  taken  as  any  value  other 
than  zero,  and  then  dy  will  be  the  perpendicular  distance 
from  the  end  of  dx  to  the  tangent  and  ds  will  be  the  distance 
from  the  point  (x,  y)  along  the  tangent  to  end  of  dy.  From 
figure,  (ds)2  =  (dx)2  +  (dy)2. 

11.   Rate,  Slope,  and  Velocity.  —  The  differential  triangle 

PDT  in  figure  for  (d)  Art,  10,  gives  —-  =  tan  <f>  =  slope  of  the 

dy 
curve  y=  f(x)  at  point  (x,  y),  and    -j-  is  the  ratio  of  the 

change  of  y  to  the  change  of  x  at  the  point  (x,  y) ,  or  for  any 

corresponding  values  of  x  and  y;  and  ~  is  called  the  rate 

of  y  with  respect  to  x. 

Aw 

-r—  is  the  average  slope  or  the  average  rate  of  change  of  y 

with  respect  to  x,  while  the  point  (x,  y)  moves  over  As  on 

the  curve  or  while  x  and  y  take  successive  values  over  any 

range. 

If  s  =f(i),  where  s  denotes  distance  from  some  origin, 

ds 
and  t,  the  time  elapsed,  then  -j-  is  the  rate  of  change  of  s 

with  respect  to  t,  what  is  called  velocity,  speed,  or  rate  of 

ds 
motion:  v  =  -=-• 
at 

In  the  case  of  uniform  motion  in  a  straight  or  curved  path, 

s      As      ds 
0  =  7  =  tt  =  TT=a  constant.     In  the  case  of  non-uniform 
t      At      dt 

ds 
or  variable  motion,  v  =  ^ •  —  a  variable. 


RATE,   SPEED,   AND   ACCELERATION  19 

In  figure  for  (d)  Art.  10,  it  is  seen  that  (ds)2  =  (dx)2  +  (dy)2; 
dividing  byW,     (fH$)V(g)'; 
.*.     velocity  of  a  point  in  its  path  is  resultant  velocity, 


ds      .  ffdxy 

frhi\- 

+  ($-<•*  +  *, 

dx 
x-component  is  vx  =  ~rr  = 

=  velocity  parallel  to  x-axis, 

^/-component  is  vy  =  -~  = 

:  velocity  parallel  to  ?/-axis; 

i^rk~dy  -dy  /dx-    • 
tan<t>-dx--Tt/Tt>    " 

dy      dx 

-rr  =  ~tt  tan  0,  or  vy  =  vx  tan  0; 

.            dy      dy  /ds 

dy      ds   .     , 

^  =  ^sm  0,  or  yy  =  y  sin  0; 

dx      dx  /ds 

cos<l>  =  Ts  =  dt/dt>    •"• 

-77  =  -77  COS  0,   Or  Ps  =  H  COS  0. 

It  appears  that  -j- ,  the  rate  of  y  with  respect  to  x,  is  the'ratio 

of  the  time  rate  of  y  to  the  time  rate  of  x. 

These  expressions  for  velocity  and  their  relations  include 
the  case  in  which  the  motion  is  uniform  or  variable  along  a 
straight  line. 

12.  Rate,  Speed,  and  Acceleration.  —  Acceleration  is 
rate  of  change  of  speed  or  velocity.     Hence,  if  the  speed  is 

changing,  -=- ,  the  time  rate  of  change  of  speed,  is  called  the 

acceleration  along  the  path,  or  the  tangential  acceleration, 
and  will  be  denoted  by  at.  The  total  acceleration  a  is  equal 
to  at,  when  the  path  is  a  straight  line;  otherwise,  they  are 
not  equal.  It  is  desirable  to  distinguish  between  speed  and 
velocity.  A  body  is  in  motion  relative  to  some  other  body 
when  its  position  is  changing  with  respect  to  that  other. 


20  DIFFERENTIAL  CALCULUS 

Change  of  position  involves  change  of  distance  or  of  direction 
or  of  both  distance  and  direction.  If  a  point  moves  con- 
tinuously in  the  same  direction,  the  path  is  a  straight  line; 
if  the  direction  is  continuously  changing,  the  path  is  a  curved 
line.  The  direction  of  motion  at  any  point  of  a  curvilinear 
path  is  the  direction  of  the  tangent  at  that  point,  and  from 
one  point  to  another  the  direction  of  motion  changes  through 

the  angle  between  the  two  tan- 
gents. Thus  from  Pi  to  P2  the 
direction  changes  through  angle 
<j>.  When  the  position  of  a  point 
changes  the  displacement  takes 
place  along  some  continuous 
path,  straight  or  curved,  and  a 
certain  time  elapses.  The  rate  at  which  the  change  of  posi- 
tion takes  place  is  the  velocity  of  the  point. 

If  the  point  moves  so  that  equal  distances  are  passed  over 
in  equal  intervals  of  time,  the  motion  is  uniform  and  the 
point  has  constant  speed,  whether  the  path  is  straight  or 
curved.  If  the  direction  also  is  constant,  that  is,  if  the 
path  is  a  straight  line,  the  point  has  constant  velocity.  Thus 
there  is  uniform  motion  with  constant  speed  either  in  a 
straight  line  or  in  a  curved  line,  but  there  is  uniform  or 
constant  velocity  in  a  straight  line  only. 

The  extremity  of  either  hand  of  a  clock  moves  in  a  circular 
path  over  equal  distances  in  equal  intervals  of  time,  but  its 
direction  is  continuously  changing.  The  motion  is  uniform 
and  the  speed  constant,  but  the  velocity  is  not  constant  since 
the  direction  is  variable.  Hence,  a  body  may  move  in  a 
circle  with  constant  speed  and  yet  its  velocity  is  variable. 
In  this  case  the  acceleration  at  along  the  tangent  is  zero, 
while  the  total  acceleration  a,  the  rate  of  change  of  the 
velocity,  is  normal,  directed  towards  the  center,  and  has  a 
constant  value  depending  upon  the  speed  and  radius.  (This 
value  will  be  derived  later.) 


ILLUSTRATIONS  21 

The  term  speed  thus  denotes  the  magnitude  of  a  velocity. 
However,  the  term  velocity  itself  is  ordinarily  used  in  the 
sense  of  speed  as  well  as  in  the  strict  sense  of  speed  and 
direction.  In  the  great  majority  of  cases  the  direction  is 
assumed  to  be  known,  and  the  magnitude  of  the  velocity  is 
what  is  in  question. 

Note.  —  A  velocity  having  both  magnitude  and  direction 
is  what  is  called  a  vector  quantity  and  can  be  represented  by 
a  straight  line  having  the  direction  of  the  velocity  and  a 
length  denoting  its  magnitude.  Hence  the  sides  of  the  tri- 
angle PTD  in  figure  for  (d)  Art.  10,  may  be  taken  to  repre- 
sent the  resultant  velocity  v  and  its  components  vx  and  vy. 

13.  Rate  and  Flexion.  —  Flexion  has  been  adopted  by 
some  writers  as  a  term  for  the  rate  of  change  of  slope.  Hence, 
when  the  slope  changes,  and  it  always  does  except  for  a 

straight  line,  -r- ,  the  rate  of  change  of  the  slope  with  respect 

to  x  will  be  called  the  flexion  of  the  curve  and  will  be  denoted 
by  b,  from  the  word  bend.  When  the  velocity  and  the  slope 
are  uniform,  there  is  no  acceleration  and  no  flexion;  that  is, 

dv  A  dm 

-rr  =  0  and  -r-  =  0. 
at  ax 

14.  Illustrations.  —  Consider  the  established  equations 
of  motion: 

s  =  vt  or  v  =  -  > 

v  =  gt  =  32t(at  =  g  =  32  ft.  per  sec.  per  sec.  approximately); 
s  =  \  gt2  =  16 1\ 

When  the  motion  is  uniform  the  velocity  or  speed  is  the 
whole  distance  divided  by  the  whole  time ;  or  any  increment 
of  the  distance  divided  by  the  corresponding  increment  of 
the  time  is  the  velocity  at  any  point,  and  it  is  the  same  as 
at  any  other  point,  since  it  is  constant. 

s      As      ds 
.'.     v  =  -  =  -r-=^-  =  Si  constant. 
t      At      dt 


22  DIFFERENTIAL  CALCULUS 

In  the  case  of  variable  motion  the  whole  distance  divided 

by  the  whole  time  gives  the  average  velocity  over  the  whole 

distance;    or  any  increment  of  the  distance  divided  by  the 

corresponding    increment    of    the    time    gives    the    average 

velocity  over  that  increment  of  the  distance.     The  velocity 

at  any  point  is  now  given  by  the  distance  that  would  be  gone 

over  in  any  time  divided  by  that  time,  if  at  the  point  the 

motion    became    and    continued    uniform    or    the   velocity 

became  constant. 

ds 
Thus    v  =  -t  =  32  t  gives  v  =  32  ft.  per  sec.  at  the  end 

of  the  first  second ;  and  means  that  the  distance  in  the  next 
second  would  be  32  ft.,  if  at  the  end  of  the  first  second  the 
velocity  became  constant. 

As  a  matter  of  fact,  s  =  16  t2  gives  16  feet  for  the  distance 
in  the  first  second,  and  48  feet  for  the  distance  actually 
passed  over  in  the  next  second.  This  variation  of  distance 
is  of  course  due  to  the  velocity  being  constantly  accelerated. 

So  when  it  is  said  that  a  train  at  any  point  is  moving  at 
sixty  miles  per  hour,  it  is  not  asserted  that  it  will  actually  go 
sixty  miles  in  the  next  hour;  but  what  is  implied  is,  that  the 
train  would  go  sixty  miles  in  any  hour  if  from  that  point  it 
continued  to  move  with  unchanged  velocity. 

Therefore,  in  ordinary  language,  variable  velocity  is  ex- 
pressed by  the  differential  of  the  distance  divided  by  the 

ds 
differential  of  the  time;  that  is,  by  -tt • 

In  the  case  above, 

ds  _  60  miles  _  1  mile  _  88  feet . 
dt  ~    1  hour        1  min.        1  sec. 

thus  dt  may  be  taken  as  any  value  other  than  zero,  if  the 
corresponding  value  of  ds  is  taken. 


EXERCISE  II  23 


EXERCISE  II. 

1.  u  =  2  x.  Show  graphically  the  change  in  u  when  x  is  given  an 
increment,  by  taking  x  as  the  base  of  a  variable  rectangle  of  altitude  2, 
and  u  as  the  area.     Is  the  change  uniform  for  u  ? 

2.  u  =  x2.  (a)  Show  graphically  the  change  in  u  when  x  is  given  an 
increment,  by  taking  x  as  the  side  of  a  variable  square,  and  u  as  the 
area.  Show  graphically  the  change  in  u  if  the  change  became  uniform. 
(6)  Show  same  when  x  is  taken  as  the  base  of  a  variable  right  triangle  of 
altitude  2  x,  and  u  as  the  area.  Show  the  change  in  u  if  the  change 
became  uniform. 

3.  V  =  x3.  Show  graphically  the  change  in  V  when  x  is  given  an 
increment,  if  the  change  in  V  became  uniform;  x  being  the  side  of  a 
variable  cube,  and  V  the  volume  of  the  cube. 

4.  If  a  body  is  moving  with  uniform  velocity  and  passes  over  1000 
feet  in  10  seconds,  what  is  its  velocity  at  any  point?  If  distance  is 
taken  as  axis  of  ordinates  and  time  as  axis  of  abscissas,  what  would  the 
slope  of  the  graph  be  ? 

5.  s  =  16 12.     Compute  the  values  of  s  when  t  =  1,  2,  1.1, 1.01, 1.001. 
Get  the  average  velocity  between  t  =  1  and  t  =  2,  between  t  =  1 

and  t  =  1.1,  between  t  =  1  and  t  =  1.01,  between  t  =  1  and  t  =  1.001. 
From  v  =  32  t,  get  the  velocity  at  t  =  1  and  compare  average 
velocities  with  it.  What  would  be  the  distance  passed  over  in  the 
second  second,  if  at  the  end  of  the  first  the  velocity  became  uniform? 
What  is  the  actual  distance  passed  over  in  the  second  second  ?  Which 
is  the  increment?     Which  is  the  differential? 

6.  If  a  ship  is  sailing  northeast  at  10  knots,  what  is  its  northerly 
rate  of  motion?     What  is  its  easterly  rate? 

If  it  is  sailing  S.  30°  W.  at  10  knots,  what  is  its  southerly  rate  ?  What 
is  its  westerly  rate? 

7.  If  the  grade  of  a  road  is  such  that  the  rise  is  52.8  feet  in  every 
mile,  what  is  the  slope? 

8.  If  the  grade  of  a  road  is  continuously  changing,  the  average  slope 
is  given  by  what  ?     The  slope  at  any  point  would  be  the  slope  of  what  ? 


CHAPTER  II. 

DIFFERENTIATION.     DERIVATIVES.     LIMITS. 

15.  Derivative.  —  The   ratio   of   the    differential   of   a 

function  of  a  single  variable  to  that  of  the  variable  is  called 

dv 
the  derivative  of  the  function.    Thus  -/  denotes  the  deriva- 

dx 

tive  of  y  as  a  function  of  x.    Since  the  derivative  may  vary 

with  x,  as  the  slope  of  a  curve  varies  from  point  to  point,  it 

is,  in  general,  itself  a  function  of  x;  hence,  the  derivative  of 

f(x)  is  appropriately  denoted  by  /'(#),  and  is  often  called 

the  derived  function.     So 

.*.    dy  =  f'(x)dx. 

Since  dy  =  /'  (#)  dx,  the  derivative  is  also  called  the  differ- 
ent/ 
ential  coefficient.     The  derivative  ~r  is  sometimes  denoted 

u     n  dx 

by  Dx2/. 

In  the  case  of  a  curve  the  derivative  is  the  slope,  in  the  case 
of  motion  it  is  the  velocity,  speed,  or  rate  of  motion;  m  every 
case  it  is  the  rate  of  change  of  the  function  with  respect  to  the 
argument  or  variable. 

Examples.  — 


dy 
erivauve 

24 


If  V  —  f  (x)  =  mx  +  b,  the  derivative  -p  =  /  (x)  =  m. 


LIMITS  25 

ds 
If  s  =  /  (0  =  vt>  the  derivative   37  =  /'  (0  =  v- 

If  y  =  /(*)  =  0*,  the  derivative  ^  =  f  (t)  =  g. 

Here  m.,  v,  and  g  are  constants. 

16.  Differentiation.  —  The  operation  of  finding  the 
differential  of  the  function  in  terms  of  the  differential  of  the 
argument,  or  the  equivalent  operation  of  finding  the  deriva- 
tive, is  called  differentiation.  The  sign  of  differentiation  is 
the  letter  d;  thus  d  in  the  expression  d  (x2)  indicates  the 

operation  of  finding  the  differential  of  x2,  and  in -7-  (x2),  that  of 

finding  the  derivative.     Dxy,  -j- ,  and  f  (x)  each  denote  the 

derivative  of  y  as  a  function  of  x. 

The  general  method  of  getting  the  derivative  of  y  =  /  (x) 

is  by  finding  the  limit  of  the  ratio  of  the  increments  of  y  and 

x  as  they  are  diminished  towards  zero  as  a  limit;    for  the 

limit  which  the  ratio  approaches,  when  defined  to  be  the 

dy 
derivative,  can  be  shown  to  be  ™ 

17.  Limits.  —  The  student  has  been  made  acquainted 
in  Geometry  with  the  notion  and  use  of  limits;  for  exam- 
ple, the  area  or  the  circumference  of  the  circle,  as  the 
limit  of  the  area  or  the  perimeter  of  the  inscribed  and  cir- 
cumscribed polygons,  when  the  number  of  sides  increases 
without  limit,  or  when  the  length  of  the  side  approaches  zero 
as  a  limit.  A  precise  statement  of  a  limit  as  used  in  the 
Calculus  is  as  follows : 

When  the  difference  between  a  variable  and  a  constant  becomes 
and  remains  less,  in  absolute  value,  than  any  assigned  positive 
quantity,  however  small,  then  the  constant  is  the  limit  of  the 
variable. 

If  x  is  the  variable  and  a  is  the  limit,  the  notation  is 
lim  x  =  a,  or  x  =  a,  or  lim  (a  —  x)  =  0,  or  (a  —  x)  —  0; 


26  DIFFERENTIAL  CALCULUS 

in  which  =  is  the  symbol  for  approaches  as  a  limit.  When 
the  limit  of  a  variable  is  zero,  the  variable  is  an  infinitesimal. 
The  difference  between  any  variable  and  its  limit  is  always 
an  infinitesimal. 

18.  Theorems  of  Limits.  —  The  elementary  theorems  of 
limits  are: 

1.  If  two  variables  are  equal,  their  limits  are  equal. 

2.  The  limit  of  the  sum  or  product  of  a  constant  and  a 
variable  is  the  sum  or  product  of  the  constant  and  the  limit 
of  the  variable. 

3.  The  limit  of  the  variable  sum  or  product  of  two  or 
more  variables  is  the  sum  or  product  of  their  limits. 

4.  The  limit  of  the  variable  quotient  of  two  variables  is 
the  quotient  of  their  limits,  except  when  the  limit  of  the  divisor 
is  zero.     (See  Note,  Art.  20,  for  proof.) 

Note.  —  The  Differential  Calculus  solves  such  limits  as 
the  exceptional  case  just  stated. 

19.  Derivative  as  a  Limit.  —  The  limit  of  a  variable,  as 
z,  is  often  written  It  (z). 

Lim    t^   ,    or   It-^-,   denotes   It(-r^)  when   Ax  =  0. 
Az=o  LAzJ  Ax'  \AxJ 

dv  A?/ 

In  defining  -^  as  a  rate  (Art.  11),  it  is  stated  that  -^  is  the 

average  slope,  or  average  rate  of  change  of  y  with  respect  to 
x,  over  the  range  Ax.     As  has  been  given  (Art.  7), 

Ay  =  Af(x)  =  f(x  +  Ax)-f(x)^ 
Ax        Ax  Ax 

where  y  =  f  (x)  and  x  is  any  value  of  x.  It  remains  to  be 
shown  that 

lim  r^i  =  lim/(*+A*)-/w = & 

Ax-oLAsJ       ax=o  Ax  dx 

(a)  By  rates  voithout  the  aid  of  a  locus.  Let  time  rates  be 
used  and  let  t,  x,  and  y  denote  any  corresponding  values  of 


REMARKS.     FUNCTION  OF  A  FUNCTION  27 

t,  x,  and  y,  from  which  At,  Ax,  and  A?/  are  reckoned.     Since 
-rj  is  the  average  rate  of  y  over  interval  Ay,  -~  is  the  time 

rate  of  y  at  a  value  of  y  between  y  and  y  +  Ay; 

Ay  _  ( time  rate  of  ?/  at  the  value  of  y  {         m 
A£  —  (     from  which  Ay  is  reckoned.    $ 

Ax  _  ( time  rate  of  x  at  the  value  of  x  )  ,  . 

A£  ~"  )     from  which  Ax  is  reckoned.  S 


Dividing  (1)  by  (2),  there  results, 

Ax=o  \_Ax\      the  time-rate  of  x      dt  j  dt      dx 

(Compare  Art.  11.) 


_  the  time-rate  of  y  _  dy  /dx  _  dy 


Thus  in  showing  the  derivative  as  a  limit,  it  appears  that  the 
derivative  of  a  function  expresses  the  ratio  of  the  rate  of 
change  of  the  function  to  that  of  its  variable.  It  is  evident 
that  a  function  is  an  increasing  or  a  decreasing  function 
according  as  its  derivative  is  positive  or  negative. 

In  the  above  derivation  in  place  of  time-rates,  the  rate  of 
any  other  variable  of  which  x  and  y  are  functions  could  be 
used. 

Remarks.       Function   of   a   Function.  —  It    should   be 

noted  that  x  and  y  being  taken  as  functions  of  a  third 

variable  t,  to  every  value  of  this  auxiliary  variable  there 

corresponds  a  pair  of  values  of  x  and  y,  so  y  is  indirectly 

determined  as  a  function  of  x.     The  derivative  of  y  as  a 

function  of  x  mediately  through  t  is,  as  shown; 

(fy  _  dy   I  dx 

dx       dt  J   dt 

a  t   .  dy      dy    dx 

Solving  gives  Tt=  Tx' Hi 

and  this  gives  the  formula  for  the  derivative  of  the  func- 
tion of  a  function.  For  if  y  is  directly  given  as  a 
function  of  x,  and  x  as  a  function  of  t,  then  y  is  said  to 


28  DIFFERENTIAL  CALCULUS 

be  a  function  of  a  function  of  t,  as  it  is  given  as  a 
function  of  t  mediately  through  x.  If  y  is  a  given  function 
of  z,  and  z  a  given  function  of  x,  then  y  is  a  function  of  a 
function  of  x,  and  the  formula  for  the  derivative  of  y  is 

dy  _  dy     dz 
dx      dz     dx 

Functions  of  functions  often  occur  and  there  may  be 
several  intermediate  variables  such  as  z  in  above  case.  A 
function,  as/  (x),  is  defined  to  be  continuous  for  the  value  a 
of  x,  or, .more  simply,  continuous  at  a,  if  /  (a)  is  a  definite 
finite  number,  and  if  lim  f(x)  =  f  (a) ;  that  is,  if  lim  /  (x)  = 

x=a 

f  (lim  x) .  By  this  definition  the  elementary  functions  of  a 
single  variable  are  continuous  for  all  values  of  the  variable 
except  those  for  which  a  function  becomes  infinite. 

A  concrete  case  in  everyday  experience  of  a  function  of  a 
function  is  the  change  in  length  of  a  metal  bar  as  the  tem- 
perature changes  with  time.  Here  the  length  is  a  function 
of  the  temperature,  and  the  temperature  is  a  function  of  the 
time;  hence  the  length  is  a  function  of  a  function  of  the  time. 
The  length,  being  directly  a  function  of  the  temperature,  is 
indirectly  a  function  of  the  time  through  the  temperature, 
which  is  directly  a  function  of  the  time. 

The  rate  of  change  of  length  per  second  is  equal  to  the 
product  of  the  rate  of  change  of  length  per  degree  and  the 
rate  of  change  of  temperature  per  second.  If  /,  T,  and  t 
denote  the  length,  temperature,  and  time,  respectively,  then, 
in  accordance  with  the  formula, 

<ti=dl     <W 
dt      dT*  dt' 

If  the  length  and  temperature  are  taken  as  changing  each 
directly  with  the  time,  then  the  rate  of  change  of  the  length 
per  degree  is  equal  to  the  rate  of  change  of  length  per  second 
divided  by  the  rate  of  change  of  temperature  per  second. 


REMARKS.     FUNCTION  OF  A  FUNCTION 


29 


The  formula  would  be 


d^ 
dT 


(R  /dT 
dtl   dt1 


which  may  be  obtained  from  the  other  formula  by  solving; 
or  the  first  may  be  obtained  from  this.  As  all  variables 
change  with  time,  that  is,  are  functions  of  time,  time  rates 
are  most  common. 

(6)   To  show  geometrically 


lim    -£=-£  =  slope  of 
Az=o  |_A*J       dx 


curve. 


Let  OPn  be  the  locus  of  y  =  f(x),  PPiS  a  secant,  and  PT 
a  tangent  at  P.     If  arc  OP  =  s,  arc  PPi  =  As.     Let 


OM  =  x,    MMi  =  Ax,     then    MP 
Hence 


DP,  =  Ay. 


Ay 
Ax 


DPi 
PD 


slope  of  secant  PPiS. 


slope  of  the  curve  y  =  /  (x)  at 


Conceive  the  secant  PPiS  to  be  re- 
volved about  P  so  that  arc  PP\ 
(=  As)  =  0;  then  Ax  =  0,  Ay  =  0,  and 
the  slope  of  the  secant  =  the  slope 
of  the  tangent  at  P. 

Hence    lim  \-~    =  -J^ 
A2=o  \_±xj      dx 

point  (x,  y). 

The  limit  is  'thus  shown  to  be  the  derivative,  whether-^ 

dx 

is  regarded  as  only  a  symbol  for  the  limit  of  the  ratio  of  the 

increments  of  y  and  x,  or  as  that  limit  and  also  a  definite 

ratio  itself  of  the  differentials  of  y  and  x. 

Corollary.  —  If  when  Ax  =  0,    -r^-   varies,   the  locus   of 

Ax 

Ay 
y  =  /  Or)  is  a  curved  line;   otherwise,  if  -~  is  constant,  the 

Ax 


30  DIFFERENTIAL  CALCULUS 

locus  is  a  straight  line  coincident  with  the  tangent,  and 

-~  =  -~ .  So  for  a  straight  line,  the  ratio  of  the  increments 
Ax      dx 

of  y  and  x,  being  constant,  does  not  approach  a  limit  as  Ax 

approaches  zero,  and  the  derivative  is  the  constant  ratio 

Ay  =  dy 

Ax     dx 

Ay 
In  general,  -~  will  approach  a  finite  limit  except  where  the 

locus  is  perpendicular  or  parallel  to  the  x-axis,  when  the 
slope  is  infinite  or  zero.  On  special  curves  where  there  are 
two  tangents  at  a  point,  the  limit  is  not  definite.  (See  Note, 
Art.  80.) 

Note.  —  This  definition  of  the  derivative  as  the  limit  of 
the  ratio  of  the  increments  of  y  and  x  as  they  converge 
towards  zero  is  the  fundamental  conception  of  the  Differential 
Calculus  by  the  method  of  limits. 

In  this  method  since  Ay  and  Ax  are  variables  each  approach- 
ing zero  as  a  limit,  they  are  infinitesimals,  for  any  variable 
with  zero  as  a  limit  is  defined  to  be  an  infinitesimal.  If  dx 
is  taken  as  always  equal  to  Ax,  then,  except  when  the  locus 
is  a  straight  line,  dy  will  alwaijs  differ  from  Ay;  and  dx,  dy, 
Ay  and  (Ay  —  dy)  are  infinitesimals  when  Ax  approaches 
zero,  for  they  each  approach  zero  as  Ax  approaches  zero. 
However,  dx  may  be  taken  as  any  increment  of  x  and,  when 
x  is  the  independent  variable,  may  be  made  a  finite  constant, 
for  x  may  be  considered  as  changing  uniformly  by  finite 
increments;  but  then,  except  for  a  straight  line,  dy  is  variable 
though  finite.  When  Ax  is  infinitesimal  any  particular  value 
of  Ax  may  be  taken  as  constant,  for  any  particular  val  ue  of 
an  infinitesimal  is  a  fixed  finite  quantity,  small  or  large  as  the 
case  may  be.  Whether  dy  and  dx  are  infinitesimals  or  finite 
quantities,  their  ratio  for  any  particular  value  of  the  variable 
is,  in  general,  constant;  and  it  is  their  ratio  that  is  important. 

Both  ways  of  regarding  differentials  are  useful.     Finite 


ILLUSTRATIVE  EXAMPLES  31 

differentials  are  desirable  for  their  simplicity,  especially  to 
make  the  differential  of  the  independent  variable  constant. 
But  when  Integration  is  regarded  as  finding  the  limit  of  a 
sum,  as  will  be  shown  later,  differentials  are  necessarily 
infinitesimal. 

One  advantage  in  making  dx  infinitesimal  and  taking  it 
very  small  is  that  Ay  is  then  very  nearly  equal  to  dy,  and  so 
instead  of  computing  Ay  in  some  investigation  the  simpler 
and  easily  found  dy  may  be  taken  for  it.  In  practical  work 
dx  and  dy  are  usually  taken  very  small  quantities,  but  it  is 
their  ratio  that  is  of  importance.  In  this  connection  it  should 
be  borne  in  mind  that,  however  small  a  quantity  may  be, 
it  is  not  an  infinitesimal  as  defined  in  the  Calculus,  unless 
it  is  a  variable  approaching  zero  as  a  limit. 

20.  Illustrative  Examples.  —  In  these  examples,  as 
elsewhere,  the  letter  symbol  for  the  argument  in  general  is 
used  as  the  symbol  also  for  some  particular  value  of  the 
argument;  this  double  use  of  the  symbol  making  for  con- 
ciseness and  generality. 

Example  1.  —  Let  the  function  to  be  differentiated  be 

y  =  x\  _2  (l) 

y  +  Ay  =  (x  +  Ax)2  =  x2  +  2x  Ax  +  Ax,  (2) 

Ay  =  2  xAx  +  Ax,  (3) 

^  =  2x  +  Ax,  (4) 

m  -*  =  lim  Rgl  =  lim  (2  s  +  Ax)  =  2x;  (5) 

ax     ^x=o\_iixj     &x=o 

.".     dy  =  2  x  dx.  (6) 

The  actual  change  of  y  corresponding  to  any  change  of  x  is 
given  by  (3).  The  average  rate  of  change  of  y  from  any 
value  of  x  to  x  +  Ax,  or  the  average  slope  of  the  curve  over 
that  range,  is  given  by  (4).  The  rate  of  change  of  y  with 
respect  to  x  at  any  value  of  x,  or  the  slope  of  the  curve  at 


32  DIFFERENTIAL  CALCULUS 

any  point  (x,  y),  is  given  by  (5).  What  would  be  the  change 
of  y  for  any  change  of  x,  if  at  any  value  of  x  the  change  of  y 
became  uniform,  is  given  by  (6) ;  and  it  shows  that,  at  any 
point  (x,  y),  y  is  changing  2  x  times  as  fast  as  x  is  changing. 
Example  2.  —  Let  the  function  to  be  differentiated  be 

s=16Z2.  (1) 

s  +  As  =  16  (t  +  AO2  =  16  (t2  +  2 1  A*  +  A?),         (2) 

As  =  32 1 M  +  16  A?,  (3) 

(4) 

(5) 
(6) 

(5) 
(7) 
(8) 

(9) 

dv      Av      00  ,    . 

a  -  a'  =  It  -  At  =  32'  (10) 

The  distance  s  passed  over  by  a  body  falling  from  rest  in  any 
time  t  is  given  by  (1).  The  actual  distance  passed  over  in 
time  A*,  after  any  time  t,  is  given  by  (3).  The  average  time- 
rate  of  the  distance,  or  the  average  velocity  from  s  to  s  +  As, 
is  given  by  (4).  The  .time-rate  of  the  distance,  or  the 
velocity  at  end  of  any  time  t,  is  given  by  (5).  What  would 
be  the  distance  passed  over  in  time  dt  (=  A*),  if  at  end  of 
time  t  the  body  moved  on  with  unchanged  velocity,  is  given 
by  (6).  The  actual  change  of  the  velocity  in  time  A*  is  given 
by  (8).  That  the  velocity  is  changing  uniformly  is  shown 
by  (9),  since  the  ratio  of  the  two  increments  is  constant. 


As 
AZ 

=  32 1  +  16  M, 

V 

ds 
~dt~ 

32*; 

.'.     ds 

=  32 1  dt. 

Let  the  function  to  be  differentiated  be 

v  = 

32*. 

v  +  Av  = 

:  32  (t  +  AO, 

Av  = 

--  32  A*, 

Av 
AZ  ~ 

=  32, 

ILLUSTRATIVE   EXAMPLES  33 

The  rate  of  change  of  the  velocity  or  speed,  the  acceleration 
a  or  a<,  is  given  and  shown  to  be  constant  by  (10). 

It  is  to  be  noted  that  — -,  being  a  constant  ratio  by  (9), 

does  not  approach  a  limit;  hence,  the  derivative  -r-  is  equal 

at 
Iv 
to  —  and  is  therefore  constant  acceleration.     (See  Corollary, 

(b),  Art.  19.) 
Note.  —  If  6  =  16  t2  be  represented  graphically  by  a  curve, 

ds 

then  the  slope  of  the  curve  is  m  =  -j;  =  32 1  —  v.  the  veloc- 

dt 

ity;  and  the  flexion  of  the  curve  is  b  =  -=■-  =  —  =  32  =  at, 

a  i       ah 

the  acceleration. 

Example  3.  —  Let  the  function  to  be  differentiated  be: 

y  =  mx  +  b.  (1) 

y  +  \y  =  m  (x  +  Ax)  +  b  =  mx  +  m  •  \x  +  b,        (2) 
ly  =  m  •  ±x,  (3) 

/.     a7?/  =  m  c2x.  (6) 

Here  again  the  ratio  of  the  increments,  being  shown  by  (4) 
to  be  a  constant  m,  does  not  approach  a  limit;    hence,  as 

shown  by   (5)   the  derivative  -f-  =  -r^-  =  m,   the   constant 

ax      Ax 

slope  of  the  line  y  =  mx  -\-b. 

That  the  ordinate  is  changing  m  times  as  fast  as  the 
abscissa  is  shown  by  (6). 

It  is  evident  that  for  a  linear  function  not  only  is  the  ratio 
of  the  increments  the  derivative,  but  the  increments  are  the 
differentials  as  defined. 


34 


DIFFERENTIAL  CALCULUS 


Example  4.  —  Let  the  function  to  be  differentiated  be 


y  =  -    or    xy=l. 

X 


(i) 

(2) 
(3) 


(x  +  Ax)  (y  +  Ay)=xy  +  xAy  +  yAx  +  Ax  Ay  =  1, 
xA2/  +  j/Ax  +  AxAj/=  0,    or    (x  +  Ax)  At/  =  -y  Ax, 

—^  = ^-r— ,  average  slope  over  Ax,    (4) 

Ax  x  +  Ax' 

cj  x 

(5) 


^  =  limr^l  =  lim(- 

dx        Ax=0  L^^J         A^=0  V 


x  +  Aa;y 
slope  at  any  point  (x,  y) ; 


dy  =  —  -dx, 
x 


(6) 


showing;  that  the  .  of  y  is  -  times  the 

increase  x 


increase 


of  x,  at  any  point  (x,  y) 


xM 


decrease 

Example  5.  —  In  compressing  air,  if  the 
temperature  of  the  air  is  kept  con- 
stant, the  pressure  and  the  volume 
are  connected  by  the  relation  pV  = 
constant.  To  find  the  rate  of  change 
of  the  pressure  with  respect  to  the 

dip 
volume,  that  is,  the  derivative  jy- 

Let  pV  =  K,  (1)     o~vm 

(p  +  Ap)  (7  +  A  7)  =  pV+p  A  7  +  7  Ap  +  Ap A V>  K,    (2) 

pA7  +  7Ap  +  ApA7  =  0,  or  (7  +  A7)Ap  =  -pA7,    (3) 

A?  = V__  (4) 

A7         7+A7'  w 

average  rate  of  change  from  7  to  7  +  A  V,  v 

%  =  hm  r^l  =  lim  (-  vl—)  =  -  £,       (5) 

d7     af^oL^^J      at'=o\      7  +  AT/  7 

rate  of  change  for  any  corresponding  values  of  p  and  V> 


ILLUSTRATIVE  EXAMPLES  35 

.'.     dp  =  —  T?dV,  showing  that  the  .  of  pressure  is 

V  increase 

^  times  the    ,  of  volume  at  any  corresponding  values 

of  pressure  and  volume. 

Example  6.  —  Let  M  be  the  mass  of  a  body,  V  its  volume, 
and  p  its  density;  then, 

AM      M 
AV       V      p' 
the  density  at  any  point,  when  the  body  is  of  uniform 
density; 

..      AM      dM 

Iim    ~jr-==   =  -p^r  =   p, 

the  density  at  any  point,  when  the  density  varies  from  point 
to  point.     Here  when  the  body  is  not  homogeneous,    the 

density  being  variable,  -z-y.  is  the  average  density  of  the 

portion  of  mass,  AM;   while  the  derivative,  -^,  expresses 

the  density  at  a  point  of  the  body  whether  the  density  is 
variable  or  uniform. 

Note.  —  In  regard  to  lim  \-r^  \  =  -j- ,  the  derivative  of  y 

as  a  function  of  x,  it  is  important  to  note  that,  since  the 
limit  of  the  divisor  is  zero,  it  is  wrong  to  write 

lim  Ay      0 


iX=0  LAzj     in 


ax=o  I  Ax  |      lim  Ax      0 


This  case  is  specially  excepted  in  Theorem  4,  Art.  18.     To 
prove  the  Theorem  4,  Art.  18: 

Since  y  =  -  •  x, 

x 

lim  y  =r  lim  -  •  lim  x,  by  Theorem  3, 

,.     y      lim?v    ..,. 
\     lim  -  =  r-. — - ,  if  lim  x  is  not  zero, 
x      lima; 


36  DIFFERENTIAL  CALCULUS 

When  lim  x  is  zero,  division  by  it  is  inadmissible  by  the  laws 

v 
of  Algebra.     If  lim  x  were  zero  and  lim  y  not  zero,  then  - 

x 

is  infinite  and  has  no  limit;  hence  the  exception  in  Theorem 
4.     The  notation  lim  -  =  oo ,  if  so  written,  means  that,  as  x 

x=Q  X 

approaches  zero  as  a  limit,  -  increases  without  limit;  that  is, 

x 

the  limit  is  non-existent. 

Infinity  or  an  infinite  quantity  is  not  a  limit,  and  the 
symbol  oo  means  a  variable  increasing  without  limit. 

In  Example  4,  where  y  =  -,  -  =  —0.    Here  where  lim  x 

xxx- 

v 
is  zero  and  lim  y  is  not  zero,  -  is  infinite,  having  no  limit. 

X  "• 

In  Example  5,  the  limit  —  £  is  finite  for  finite  values  of  p 
and  V.    From  V  —  t?  and  tJ  =  ™,  p  —  °°  as  F  =  0;  hence 

as  lim  V  is  zero  and  lim  p  is  not  zero,  ^  is  infinite; 

At?  p 

•'•     A7  =  ~  y  +  AF  is  infinite  as  V  ~  °> 

[Apl 
xr=    is  non-existent  when  V  is  infinitesimal. 
A7J 

Ai/ 
In  Example  3,  where  y  =  mx  -\-b,  Ay  =  mAa;,  and  ~-  =  m. 

If  Ax  =  0,  Ay  =  0,  but  their  ratio  is  constant  and  ap- 
proaches no  limit.  Since  Ay  =  mAx,  the  law  of  change  of 
the  variables  is  known  and  the  ratio  of  two  infinitesimals  is 
a  finite  constant. 

In  Example  1,  where-  y  =  x2, 

Ay  =  2xAx  +  Atf,    ^  =  2z  +  A:r,     .%     lim  T^l  =  2x. 

Here  the  limit  of  the  ratio  of  two  infinitesimals  is  a  finite 
constant  for  any  particular  finite  value  of  x;  but,  as  x  may 


LIMIT   OF   INFINITESIMAL  ARC   AND   CHORD        37 

have  any  value,  the  limit  of  the  ratio  may  be  zero,  finite,  or 
non-existent. 

Thus  it  is  seen  that,  no  matter  how  small  two  quantities 
may  be,  their  ratio  may  be  either  small  or  large;  and  that,  if 
the  two  quantities  are  variables  both  with  zero  as  their  limit, 
the  limit  of  their  ratio  may  be  either  finite,  zero,  or  non- 
existent, but  is  not  0  0.     (See  Art.  219.) 

To  find  lim  U-^  ,  as  in  the  illustrative  examples,  the  limit 
Ax-o  LA x\ 

of  an  equal  variable  is  found;  which  limit  is,  in  general, 
determinate  and  not  identical  with  the  indeterminate  expres- 
sion 0  0.  In  certain  cases  the  limit  of  the  ratio  of  two 
infinitesimals  is  found  by  finding  the  limit  of  some  other 
variable  which,  though  not  equal  to  the  ratio,  has  the  same 
limit.  Examples  of  such  cases  will  be  given  further  on. 
(See  Art,  70,  Art.  77.) 

21.  Replacement  Theorem.  —  The  limit  of  the  ratio  of 
two  variables  is  the  same  when  either  variable  is  replaced  by  any 
other  variable  the  limit  of  whose  ratio  to  the  one  replaced  is  unity. 

Let  0,  0i,  0,  and  fa,  be  any  four  variables,  so  that 

lt^  =  \,        U4-=1,     and     ltS-  =  c.  (1) 

0i  fa  <t> 

0       6     6\     4>\  _  6\     0      fa . 
<t>       0      01      01        fa     0i       <t> 

...      ul  =  H^.lte-.lt*l  =  lt^,  by(l) 

0  fa  01  0  01 

in  which  0  is  replaced  by  0i,  and  0  by  fa,  but  the  limit  of  the 
two  variables  is  the  same. 

22.  Limit  of  Infinitesimal  Arc  and  Chord.  —  The  limit  of 
the  ratio  of  an  infinitesimal  arc  of  any  plane  curve  to  its  chord 
is  unity. 

Since  s  (Art,  19  (6),  figure)  is  a  function  of  x, 


38  DIFFERENTIAL  CALCULUS 

But        It chord  PPl  =  u  sec  DPP,  =  sec  DPT  =  ^-         (2) 
Ax  dx 

Dividing  (1)  by  (2),  Mm  [^_]  =  l.  (3) 

It  follows  from  Art.  21  that  in  a  limit  an  infinitesimal  arc 
may  be  replaced  by  its  chord. 

ALGEBRAIC  FUNCTIONS. 

23.   Formulas  and  Rules  for  Differentiation.  —  By  the 

general  method  any  function  can  be  differentiated,  but  it  is 
usually  more  directly  done  by  formulas  or  rules  established 
by  the  general  method  or  by  other  methods. 

In  the  following  formulas  u,  v,  y,  and  z  denote  variable 
quantities,  functions  of  x;   and  a,  c,  and  n,  constant  quan- 

"  d  " 

tities.     If  in  the  formulas     ~r     or  "Dj"  be  substituted  for 

ax 

"d"  and  in  the  rules  " derivative"  be  substituted  for  differ- 
ential, they  are  still  valid. 

[I]  If  y  =  x,  dy  =  dx. 

The  differentials  of  equals  are  equal. 

[II]  d  (a)  =  0. 
The  differential  of  a  constant  is  zero. 

[III]  d  (v  +  y  +  •  •  •  -  z  +  c)  =  dv  +  dy  +  •  •  •  -  dz. 
The  differential  of  a  polynomial  is  the  sum  of  the  differentials 

of  its  terms. 

[IV]  d  (ax)  =  a  dx. 

The  differential  of  the  product  of  a  constant  and  a  variable  is 
the  product  of  the  constant  and  the  differential  of  the  variable. 
[Vo]  d  (uy)  =  ydu-\-  udy. 

The  differential  of  the  product  of  two  variables  is  the  sum  of 
the  products  of  each  variable  by  the  differential  of  the  other. 
[V6]    d  (uyz  .  .  .  )  =  (yz  .  .  .  )  du  +  (uz  .  .  .  )  dy 
+  (uy  .  .  .  )  dz  +   •  •  •  • 


DERIVATION  OF   [I] 


39 


The  differential  of  the  product  of  any  number  of  variables 
is  the  sum  of  the  products  of  the  differential  of  each  by  all  the 
rest. 

Y\       D-dN-N-  dD 


[VI] 


D 


D2 


the  differential  of  a  fraction  is  the  denominator  by  the 
differential  of  the  numerator  minus  the  numerator  by  the 
differential  of  the  denominator,  divided  by  the  square  of  the 
denominator. 

[VII]  d(xn)  =  nx^dx. 

The  differential  of  a  variable  with  a  constant  exponent  is  the 
product  of  the  exponent  and  the  variable  with  the  exponent  less 
one  by  the  differential  of  the  variable. 

24.  Derivation  of  [I],  —  If  i/  is  continuously  equal  to  x, 
it  is  evident  that  y  and  x  must  change  at  equal  rates; 
that  is, 

dy  _  dx 
dx      dx' 

dx 

Since  -=-  =  h  the  rate  of  x  is  the  unit 
dx 

rate,  so  in  general  the  rate  of  a  variable 

with  respect  to  itself  is  unity,  or  the 

derivative  of  /  (x),  when  /  (x)  is  x,  is 

one. 

Geometrically   the   locus   of   y  =  x  is   the   straight   line 

through  origin  making  angle  <t>  =  j  with  z-axis. 

V      At/ 

tan  4>  =  -  =  -~  =  1, 

x        AX 


dy  =  dx. 


i    •        Av  .  L    Aw      dy 

and  smce  -r-2  is  constant,  -r2-  —  —-  =  1. 
Az  Az      dx 


dy  =  dx. 


For  examples  of   [I],  if  y2  =  2  px,  d(y2)  =  d(2px);    or  if 
x2  +  y2  =  a2,  d  (x2  +  y2)  =  d  (a2)  =  0. 


40  DIFFERENTIAL  CALCULUS 

25.  Derivation  of  [II].  —  By  definition  the  value  of  a 
constant  is  fixed,  therefore  the  rate  of  a  constant  is  zero; 
that  is, 

^.=  0,     .'.     da  =  0. 
•  dx 


Aw  .        Aw  . 

Aw  =  0,     .*.     i-^  =  0,  and  since  -r^  is  constant, 
-  *.  Ax  Ax 


If  w  =  a,  a  change  in  x  makes  no  change  in  w,  hence 
Ay 
Ax 

^  =  ^  =  0,     .-.     dy  =  da  =  0. 
Ax      dx        '  * 

Geometrically  the  slope  of  w  =  a  (a  line  parallel  to  z-axis) 
is  at  every  point  zero. 

26.  Derivation  of  [III].  —  It  is  manifest  that  the  rate  of 
the  sum  of  v  +  y  +  ■  ■  ■  —  2  +  c  is  equal  to  the  sum  of 
the  rates  of  its  parts,  v,  y,  .  .  .   —  z  and  c;  that  is, 

d  (v  -f  y  +  •  •  •  —  g  +  c)  _  dv      dy  _dz      dc 

dx  dx      dx  dx      dx 

Multiplying  by  dx,  since  dc  =  0,  the  result  is  [III].  The 
rule  shows  that  differentials  are  summed  like  any  other 
algebraic  quantities.     For  example, 

d  (b2x2  ±  a2y2  -  a2b2)  =  d  (b2x2)  ±  d  (a2y2)  -  d  (a2b2). 

27.  Derivation  of  [IV].  —  Since  A  (ax)  =  a  Ax,  the  ratio 
of  the  increments  is  constant   and  ax  changes  uniformly 

with  respect  to  x.     Hence  by  definition 
- — f    of  differentials  d  (ax)  =  adx.     If  y  =  ax, 

-j-  =  a,  slope   of  line.     Geometrically,  if 


dz 


o  «r         MdxM, 

z  =  ax  be  area  of  a  rectangle  of  base  x  and 
altitude  a,  then  the  rectangle  MPPiMi  is  the  change  of  z 
made  by  a  change  Ax  ( =  dx)  of  x,  and  being  a  uniform  change 
is  the  differential  of  z, 

.'.     dz  =  d  (ax)  =  a  dx. 


DERIVATION   OF   [Yt] 


41 


For  examples: 

d(2px)  =  2pdx,     and     d(~)  =  dl-x)  = —- 

[Va]  will  be  seen  to  include  [IV]  as  a  special  case. 

28.  Derivation  of  [VJ.  —  Let  2  =  uy;  then  z,  a  function 
of  u  is  a  function  of  y  also.  Geometrically,  let  u  and  1/  be 
the  base  and  altitude  of  a  variable  rectangle  conceived  as 
generated  by  the  side  y  moving 
to  the  right  and  the  upper  base 
u  moving  upward;  then  z  =  uy 
is  the  area.  If  at  the  value 
OMPN,   du  =  MJkfi,  and  dy  = 


dy 


*—4 
1 


Z=uy 


NNi,  the  differential  of  the  area      °  «  M      Mi 

is  MMiDP  +  NPBNh  as  that 

sum  would  be  the  change  of  the  area  of  the  rectangle  due  to 
the  change  of  u  and  y,  if  at  the  value  OMPN  the  change  of 
its  area  became  uniform.  Hence  dz  =  d  (uy)  =  y  du  +  udy. 
Here  Az  =  A  (uy)  =  d  (uy)  +  P  DPiB,  since  that  sum  is  the 
actual  change  of  the  area  due  to  the  change  of  u  and  y. 

It  is  to  be  noted  that  iiy  =  u,  then  the  rectangle  is  a  square 
and  area  z  =  u2, 

.'.     d  (u2)  =  udu  +  udu  =  2udu. 

If  y  =  a,  z  =  au,  dz  =  adu  +  uda  =  adu,  since  da  =  0. 
Hence  [IV]  is  a  special  case  of  [VJ. 

29.   Derivation   of    [VJ.  — To   prove   d(uyz)=yzdu  + 
uzdy  +  uy  dz.     If  in  [VJ,  yz  is  put  for  y; 

d  (uyz)  =  yzdu  +  ud  (yz) 

=  yz  du  +  u  (z  dy  +  y  dz) 
=  yzdu  +  uz  dy  +  uy  dz. 

By  repeating  this  process  the  rule  is  proved  for  any  number 

of  variables.     If  y  =  z  =  u, 

then     d  (uyz)  =  d  (u3)  =  u2  du  +  u2  du+'u2  du  =  Zu2  du. 


To  derive  d(uyz)  geometrically,  let  V  =  xyz  =  uyz. 


42 


DIFFERENTIAL  CALCULUS 


1 


ife 


*i—X 


If  x,  y,  and  z  be  the  edges  of  a  variable  right  parallelopiped 
conceived  as  generated  by  the  face  yz  moving  to  the  right, 
the  face  xz  moving  to  the  front,  and  the  face  xy  moving 

upward,  then  the  volume  is 
the    product    of    the    three 
__^yr        edges;  that  is,  V  =  xyz. 

If  at  the  value  OP,  Ax  = 
AAh  dy  =BBh  and  dz  =  CCh 
the  differential  of  the  volume 
is  PAi  +  PB1  +  PCi;  as  that 
sum  would  be  the  change  of 
\  the  volume  of  the  parallel- 

opiped due  to  the  change  of 
x,  y,  and  z,  if  at  the  value  OP 
the  change  of  its  volume  became  uniform.     Hence 

dV  =  d  (xyz)  =  yzdx  +  xz  dy  +  xy  dz. 
^        Here 

AV  =  dV  +  PNi  +  PLi  +  PMi  +  PPi, 

since  that  sum  is  the  actual  change  of  the  volume  due  to  the 
change  of  x,  y,  and  z.  If  y  =  z  —  x,  then  the  parallelopiped 
is  a  cube  and  V  =  xz, 

.*.     d  (xz)  =  x2  dx  +  x2  dx  +  x2  dx  =  3  x2  dx. 

dent),  then  zx  =  y. 

:.     d  (zx)  =  xdz  +  zdx 


Derivation  of  [VI].  —  Let  z  =  -  (x  and  y  indepen- 

x 


dy,   by[VJ. 


Solving, 


Corollary.  — 


dy      zdx 


dz  =  —  — 


di*)  =  ^ 


x 

x) _  xdy  —  ydx 

x  x2 


, /a\  a dx 


DERIVATION  OF   [VII]  43 

-  ,  /a\      xda  —  a  dx  adx 

for     d[-}=  5 = — ,    since  da  =  0. 

\x/  x-  x- 

-  ,  /x\      adx  —  x da      dx  7 

for    a   -  = ; =  —  j    since  aa  =  0 ; 

\a/  a-  a 

hence,  for  a  fraction  with  constant  denominator,  use 
[IV].  For  another  derivation  of  [VI],  see  Corollary  of  next 
Art.  31. 

31.   Derivation  of  [VII].  —  I.     When   the  exponent  is  a 
positive  integer. 

(a)    If  n  is  a  positive  integer,  xn  =  x  •  x  •  x  •  ton  factors; 
hence, 

c?  (xn)  =  d(x-x  -  x  to  n  factors) 

=  xn~l  dx  +  xn~l  dx  +  xn~l  dx  +  •  •  • 
to  n  terms,  by  [V&], 


(b)   By  the  general  method.     Let  y  =  xn. 
y  +  Ay—  (x  +  A#)n  =  xn+  7?xn_1  Ax  +  (terms 

with  common  factor  Ax"),  by 

A?/  =  nx1l~l  Ax  +  (terms  with  factor  Ax~)  [  Binomial 
Ay  „   .   ,.  ..,    „  ".   ,  Theorem. 


=  nxn~l  +  (terms  with  factor  Ax), 

-j-  =  lim  |  ~^L  |  =  nx*-1,     .'.     dy  =  nxn~l  dx. 
ax      az=o 


m 

II.    When  the  exponent  is  a  positive  fraction. 


Let  y  =  xn, 

then  yn  =   xm, 

:.     d(yn)  =  d(xm), 
nyn~l  dy  =  mxm~l  dx, 


1        m  xm  l  ,        mxm~1y 

dy  = rdx  = -. 

n  yn~l  n     yn 


44  DIFFERENTIAL  CALCULUS 


(*)-  = 


m 

j.m — 1  /v.  n  iyy\       _  J 

dx  =  —  xn     dx* 


n      xm  n 

III.    When  the  exponent  is  negative. 
Let  y  =  x~n,  n  being  integral  or  fractional;  then  y  =  — ; 

•'•  dy  =  d  fcr») =  ~  i£~ ** by  [VI]- Cor" Art- 30, 

dy  =  d  (x~n)  =  —  nx_n_1  dx. 
Corollary.  —   d  ( -  J  =  d  (xy~l)  =  y~x  dx  —  xy~2  dy 

_  dx  _xdy  _  ydx  —  xdy  ,VT, 

y       y2  y2 

Note.  —  A  general  proof  of  [VII]  by  logarithms,  given 
further  on  (Art.  37),  includes  the  case  where  the  exponent  is 
incommensurable.  So  the  Formula  or  Rule  is  valid  for  any 
constant  exponent.  It  is  called  the  Power  Formula  and  is 
of  most  frequent  application. 

Examples.  — 

d  (Vx)  =  d  (x 2)  =  jz  x~%  dx  = 


2  2Vx       • 

dl— 7-  j  =  d  (x~%)  =  —  =  x~*dx  = =• 

Wx)  2  2  Vx3 

d(xv*)=  V2xv'2-1dx(=  1.414 xAUdx,  approximately). 

d  (xT)  =  tx*-1  dx  (=  3.1416  x2Amdx,  approximately). 

d  ((ax  +  b) n)  =  n  (ax  +  6) n-1  d  (ax  -\-b)  =  na  (ax  +  b) n_1  dx, 

.*.      3-  ((ax  +  b)n)  =  na  (ax  +  6)n_1. 
ax 

Note.  —  The  last  example  may  be  seen  to  be  an  application 
of  the  formula  for  the  derivative  of  a  function  of  a  function. 
For  let  y  =  (ax  +  b)n  and  put  z  =  ax  +  b,  then  y  =  zn; 
now  y  is  a  function  of  z,  and  z  is  a  function  of  x;  that  is,  y  is 


EXERCISE  III  45 

a  function  of  a  function  of  x.     The  formula  given  in  Remarks, 

„_   .      dy      dy   dz . 

Art.  19,  is     /  =  /-^-' 
'         dx      dz    dx 

.-.     ~r  =  -r  ((ax  +  6)n)  =  -^-^  •     v    ,       J  =  nzn~l  •  a 
dx      dx  dz  dx 

=  na(ax  +  6)n_1. 

In  applying  Rule  [VII],  if  all  within  the  parenthesis,  as 
(ax  +  b) ,  is  regarded  as  the  variable,  the  actual  substitution 
of  z  may  be  dispensed  with  in  getting  the  derivative  of  such 
functions. 

EXERCISE  m. 

By  one  or  more  of  the  formulas  I-VII  differentiate: 

l.  y 


6^-2  +  _4         2       3 
Vx3      x      & 


dy  =d(6  si)  +  d  (4  x~§)  -  d  (2  x"1)  +  d  (3  x~4). 

§y=  J_         6         2       12 

dx      </x      Vtf      x2      x5' 

2.  i/  =  3X3  -  4x2  -  2. 

dy  =  d(3x3)  -d(4x2)  -  d  (2). 
dy  —  9x2dx  —  8xdx  —  0; 

|H  =  9  x2  -  8  x  =  (9  x  -  8)  x; 

that  is,  ?/  changes  (9  x  —  8)  x  times  as  fast  as  x. 

When    x  =  —  1,  y  is  increasing  at  the  rate  of  17  to  1  of  x; 

x  =  f ,  y  is  neither  increasing  nor  decreasing; 

x  =  0,  y  is  neither  increasing  nor  decreasing; 

x  =  },  ?/  is  decreasing  at  the  rate  of  $•  to  1  of  x; 

x  =  1,  2/  is  changing  at  the  same  rate  as  x; 

x  =  —  |,  y  is  changing  at  the  same  rate  as  x; 

x  =  2,  ?/  is  increasing  at  the  rate  of  20  to  1  of  x. 
Note.  —  In  this  way  the  meaning  of  each  differential  equation  may 
be  shown. 

3.  y  =  (1  +  2  x2)  (1  +  4  x3).         dy  =  4  x  (1  +  3  x  +  10  x3)  dx. 

dy  =  (1  +  2  x2)  d  (1  +  4  x3)  +  (1  +  4  x3)  d  (1  +  2  x2); 
or  dy  =  d(l +2x2 +  4x3 +  8X5). 


46  DIFFERENTIAL  CALCULUS 

4.  ?/  =  (x  +  l)5(2x-l)3.        dy  =  (16x  +  1)  (x  +  l)4  (2x  -  l)2dx. 

5.  y  =  (a+x)Va-*.  ^  =  2V^=T 

6.  ?/  =  (l-3a;2+6x4)(l+x2)3.  dy  =  60  x5  (1  +  x2)2  dx. 

/  i         i\4  dy      4:  (x*  —  a*)3 

7.  „.(x.-„.).  5°       3ai       • 

_  x  +  a2  dy       b  —  a* 

y  ~  x  +  b'-  dx      (x  +  bY 

(x  +  b)  d  (x  +  a2)  -  (x  +  a2)  d  (x  +  b) 


dy  = 


ix  +  6)2 


2x  -  1  dy  2x 

9*   y  ~  (x  -  l)2"  dx  (x  -  l)3" 

10.   y  =  x  (x3  +  5)*.  ^  =  5  (x3  +  1)  (x3  +  5)*. 


11.  y  = 


dx 

2  x4  dy  _  8  a2x3  -  4  x5 


12.   y  =  v  ox2  +  bx  +  cf 

T 

14.   y  = 


dx         (a2  -  x2)2 
dy  _         2  ax  +  6 
d*      2  Vax2  +6x  +  c 

^  =  1 

dx      (1  -  x)  Vl  -  x2 

dy  a2 


15.   y  = 


Va2  -  x2  dx       V(a2  _  X2)3 

xn  dy  _       nxn_1 


(1  +  x)n  dx       (1  +  x)7^1 

_xM-l  dx  2  nxn~l 

16-   ^  -^~T  dy  (x*-l)2' 

17.  y  =  ^=JL —        *.ir^±w+j,]. 

Va2  +  x2  -  x  dx      a-  LvV  +  x2  J 

Rationalize  the  denominator  before  differentiating. 

^  dx  — 

18.  x  =  t  (P  +  a2)  2    .  ^  =  (nf2  +  a2)  (C-  +  a2)  2  . 

19.  A  vessel  is  sailing  due  north  20  miles  per  hour.  Another  vessel, 
40  miles  north  of  the  first,  is  sailing  due  east  15  miles  per  hour.  At 
what  rate  are  they  approaching  each  other  after  one  hour?  After  2 
hours?  Ans.  Approaching  7  mi.  per  hr.;  separating  15  mi.  per  hr. 
When  will  they  cease  to  approach  each  other,  and  what  is  then  their 
distance  apart?     Ans.  After  1  hr.  16  m.  48  sec;   24  mi. 

20.  If  a  body  moves  so  that  s  =  Vt,  show  that  the  acceleration  is 
negative  and  proportional  to  the  cube  of  the  velocity.  Negative  sign 
shows  what  ? 


FORMULAS  AND  RULES  FOR   DIFFERENTIATION       47 

21.   If  x  =  at  and  y  =  bt  —  -^ct2,  find  -^  and  ~^- 
2  ax  dy 


d_y 
dx 

dy  1  dx      b  —  ct 
~  dt  1    dt           a 

dx 

dy  ' 

dx  I 'dy  _      a 
"  dt  1   dt  ~  b  -  ct 

(By  Art.  19(a).) 

Or 

bx      1  ex2 
y~~a~~2~tf'      •*' 

dx 

b       ex      b      cat 
a      a2       a       a2 

b  -ct 
a 

22. 

If  p  =  Vl  and  0  =  £2  ■ 

-10, 

d6 

dp  _dp  /dd  _    1     / 
dd~  dt  /  dt       2t*/ 

2t  = 

—3.     (By  Art.  19, 
4£f 

Remarks.) 

Or 

p2=t  =  (d  +  10)*. 

»»d»-          ** 

1 

&  p  up  — 

2  (0  +  10) 

*      ±$ 

23. 

Find 

,   The  equation  pV  =  C  expresses  Boyle's  law,  C  being  a  constant. 

^  and  ^-     (See  Ex.  5,  Art.  20.) 
iV          dp 

24.  The  heat  H  required  to  raise  a  unit  weight  of  water  from  0°  C. 
to  a  temperature  T  is  given  by  the  equation, 

H  =  T  +  0.00002  T-  +  0.0000003  T3. 

(a)  Find  ^.     (6)  Compute  the  numerical  value  of  the  rate  for 

T  =  35°.  Ans.  (6)  1.0025025. 

25.  A  vessel  in  the  form  of  an  inverted  circular  cone  of  semi-vertical 
angle  30°,  is  being  filled  with  water  at  the  uniform  rate  of  one  cubic  foot 
per  minute  At  what  rate  is  the  surface  of  the  water  rising  when  the 
depth  is  6  inches?    When  1  foot?    When  2  feet? 

Ans.  0.76  in.;  0.19  in.;  0.05  in.,  per  sec. 

26.  Show  that  the  slope  of  the  tangent  to  the  curve  y  =  x3  +  4  is 
never  negative.  Find  the  slope  for  x  =  0,  for  x  =  2.  For  what  values 
of  x  does  the  slope  decrease  as  x  increases? 

LOGARITHMIC  AND  EXPONENTIAL  FUNCTIONS, 

32.  Formulas  and  Rules  for  Differentiation. — 

[VIIIJ    d  (log6*)  =  —  dx  (x  positive), 
x 


(m  =  log6e  =  0.434  .  .  .  ,  for  b  =  10). 

x)  =  -  dx  (x  positive). 

(m  =  logee  =  1,  e  =  2.718  .  .  .  ). 


[VIIIJ   d  (loge  x)  =  -  dx  (x  positive). 
x 


48  DIFFERENTIAL  CALCULUS 

The  differential  of  the  logarithm  of  a  variable  is  the  product 
of  the  modulus  of  the  system  and  the  reciprocal  of  the  variable 
by  the  differential  of  the  variable. 

[IXJ  d  ibx)  =  bx loge  b  dx  (b  positive). 

[IXt]  d(ex)  =  e«dx. 

The  differential  of  an  exponential  function  with  constant  base 
and  variable  exponent  is  the  product  of  the  function  and  the 
Napierian  logarithm  of  the  base  by  the  differential  of  the 
exponent. 
[X]  d  (y*)  =  y" logey  dx  +*yK-1  dy9 

(y  positive  and  independent  of  x). 

The  differential  of  an  exponential  function  with  base  and 
exponent  variable  is  the  sum  of  the  results  obtained  by  differen- 
tiating as  though  the  base  were  constant  and  then  as  though  the 
exponent  were  constant. 
33.  Derivation  of  [VIIIa]  and  [VIII6].— 
(i)    Taking  n  an  arbitrary  constant,  let 

x  =  ny.  (1) 

log6  x  =  log6  (ny)  =  log6  y  +  log6  n.  (2) 

d  (logt  x)=d  (log*  y)[+d  (log*  n)  =  0].  (3) 

Differentiating  (1)  and  dividing  result  by  (1), 

*l  =  tyi.  (4) 

x  ""  y 

Dividing  (3)  by  (4)  gives  as  result, 

d(\ogbx)/d^  =  d(\ogby)//^.  (5) 

It  is  manifest  that  the  equal  ratios  in  (5)  are  constant  for 
any  particular  value  of  y.  Let  m  denote  the  constant  value 
of  the  ratio  when  y  —  y\\  then 

d  (log*  x)  =  —  dx,  (6) 

x 


DERIVATION  OF  [VIIIa]  AND   [VIII6]  49 

when  x  =  nyi]  and,  as  n  is- an  arbitrary  constant,  nyi  denotes 
any  positive  number.  Hence  (6)  or  [VIIIa]  holds  true  for 
all  positive  values  of  x,  m  being  a  constant.  The  constant 
m  is  called  the  modulus  of  the  system  of  logarithms,  whose 
base  is  denoted  by  b  in  this  derivation.  The  general  base  is 
often  denoted  by  a. 

The  system  whose  modulus^s  unity  is  called  the  Napierian 
or  natural  system.  The  symbol  for  the  base  of  this  system 
is  e,  called  the  Napierian  base  from  the  name  of  the  dis- 
coverer of  logarithms. 

Hence  d  (loge  x)  =  -  dx.     [VIII&] 

x 

(ii)   By  the  general  method  of  limits,     (x  positive.)     Let 

y  =  log6  x.     y  +  ky  =  logb  (x  +  Ax), 

Ay  =  log6  (x  +  Ax)  -  log&x  =  log6  (l  +  —J , 

x 
Raising  each  member  to  -r—  power  gives 

The  limit  of  each  member  as  A x  =  0  gives 

X 

v     Lz£      ..■/*.  AxV*      r     /,   .   1\»       ...       Ax      1 

lim  b  Az  =  lim  [  H =  lim  ( 1  +  -     ,  putting  — ■  =  -, 

az=o  az=o\         x  J        n=*\       n}  x       n 

so  that  (if  x  is  not  zero),  as  Ax  =  0,  n  =  <x>  ; 

x^-  x^  /  1\« 

lim 6  **  =  b  dx=  lim    1  +  -)    =e 

(denoting  the  limit  by  that  letter) ; 

...     xf^lo&e,   ort  =  !2S^  =  - 
ax  dx         x  x 

(m  =  logb  e  =  the  modulus) ; 


50  DIFFERENTIAL  CALCULUS 


Hence, 


d(\ogbx)  =  -dx.        [VIIIC 

X 


d  (loge  x)  =  -  dx,  since  loge  e  =  1  =  the  modulus.     [VIII6] 

x 

The  limit  of  ( 1  +  - )    as  n  is  increased  without  limit  is  e, 
the  Napierian  base.     (See  next  Art.  34.) 

34.   Lim  (l  +  -Y*  =  e.  (See  Ex.  7,  Art.  221.) 

™=«   \         it) 

When  n  is  a  positive  integer,  by  the  Binomial  Theorem, 

1      w(w-l)     1    ,  n(n-l)(n-2)    1 
1+n'n+     1-2     V+        1.2-3         V+'*' 

=  1  ,  1  ,    __V n)         V        n/V        n/ 

"^ i"1"      L2      "^  1-2- 3  "^  "  ' 

In  the  expansion  there  are  (n  +  1)  terms  in  all,  and  every 
term  after  the  second  can  be  written  in  the  form  given  to  the 

3rd  and  4th  terms.     As  n  =  oo ,  -  =  0, 

n 


K)' 


nj  ^     ^2^2.3^2-3.4 

=  e  =  2.7182818  .... 


Note  1.  —  The  limit  is  denoted  by  e,  which  is  an  irrational 
number,  and  was  proved  by  Hermite,  in  1874,  to  be  trans- 
cendental or  non-algebraic.  The  number  e  was  the  first 
number  to  be  proved  transcendental.  Not  until  1882  was 
the  attempt  to  prove  the  number  -k  transcendental  successful. 
This  was  finally  done  by  Lindemann.  The  proofs  consist  in 
showing  that  neither  of  the  two  numbers  is  the  root  of  an 
algebraic  equation  with  integers  for  coefficients.  Algebraic 
real  numbers  are  defined  as  those  real  numbers  which  are 
roots  of  such  an  equation.  The  importance  of  these  two 
numbers,  considered  the  most  important  in  mathematics, 


DERIVATION  OF  VALUE  OF  e  51 

warrants  some  notice.  They  are  connected  by  the  remark- 
able relation,  e*s~x  =  1.     (See  Ex.  10,  Exercise  XLIII.) 

Note  2.  —  The  above  derivation  of  the  limit  is  not  com- 
plete, for  the  result  is  true  not  only  when  n  is  "a  positive 
integer,"  but  also  for  n  positive  or  negative,  integral,  frac- 
tional, or  incommensurable.  The  value  of  the  limit,  e,  can 
be  easily  computed  to  any  desired  degree  of  precision  by 
taking  a  sufficient  number  of  terms  of  the  series.  Twelve 
terms  gives  the  result  correct  to  seven  decimals;  that  is, 
e  =  2.7182818  .  .  . 

By  comparing  the  sum  of  (n  +  1)  terms  of  the  series  with 

the  sum,  I  +  I  +  2  +  92  +  '  *  "  o^i '  wmcn  is  greater  than 

the  othor,  and  equal  to  3  —  |n_1,  it  is  manifest  that  no  matter 
how  great  n  may  be,  the  sum  of  the  (n  +  1)  terms  is  certainly 
finite  and  less  than  3.     The 

^(1+l+k+h+  ■■+£} 

may  be  considered  as  c,  or  as  usually  written, 

to  infinity.     (See  Ex.  5,  Art.  215.) 

Without  expanding,  the  lim  ( 1  +  - )    can  be  computed  to 

n  =  x  \  11/ 

any  desired  number  of  decimals  by  giving  increasing  values 
to  n;  thus, 

(1  +  -A)10  =  2.59374. 
(1.01)100  =  2.70481. 

(1.001)1000  =  2.71692. 

(1.000001)1000000  =  2.71828. 
The  last  number  agrees  with  the  value  of  e,  the  required 

limit,  to  five  decimals. 

1 

Corollary.  —  Lim  (1  +  n)n  =  e.     (See  Ex.  8,  Art.  221.) 

n=0 


52  DIFFERENTIAL  CALCULUS 

35.   Derivation  of  [IXa]  and  [IXJ.  — 
(i)    Let  y  =  b*,  then  loge  y  =  x  loge  b. 

d  (loge  y)  =  d(x  loge  6),     or    —  =  loge  b  dx, 

:.     dy  =  d  (bx)  =  bx logc b dx  (b  being  positive).     [IXa] 
Hence,   d{ex)  =  ex  dx  (since  logee  =  1).     [IX&] 

(ii)    \iy  =  bx,x  =  \oghy.     dx  =  d(\ogby)  =  1^^. 

""'     *  =  lo&e ^  =  6* l0ge 6 ^  (since  log^e  =  l0ge T    'IXa' 
Hence,  d(ex)  =  exdx.     [IX*,] 

(iii)    By  the  general  method  of  limits.     Let 

y  =  ex.        y  +  Ay  =  ex+Ax. 
Ay  =  ex+Ax  —  ex  =  ex  (eAx  -  1). 
Ay  _  ex  (eAx  —  1) 
Ax  ~         Ax 

Ax=o|_A:r_|      dx  ~  a*=o.L    \    Ax    )\~  6  AS  V    Ax    / 

=  e*  (since  lim  ^  ""  *)  =  l)     {Cor.,  Art.  36); 

.*.     dy  =  d(ex)  =  ex  dx.     [IX*] 

Corollary.  —    d  (&*)  =  bx  loge  6  dx.     [lXa] 

1  i  I 

For  if  —  =  loge  b,  6X  —  em ,  since  6  =  em : 


d(bx) 


=  d[emJ; 


:.     d(b*)  =e™  —  =  te- 
rn m 

:.     d  (bx)  =  bx  loge  b  dx. 
36.   Lim  (l  +  -V  =  e*. 

\ 


DERIVATION  OF  [X]  53 

In   limfl  +-)  ,    if   x  ^  0,    by   putting   n  =  Nx,   when 

n  =  ao  \  71/ 

n  =  oo  so  is  N  =  oo  ;  hence 

HH'+*H('+*)T 

and 

5i(1+9'  -UK1  +*)'?  =  H1 +^)T  =  e*' 

since  lim/  (z)  =  /  (lim  a;),  the  case  of  a  function  of  a  function. 
(See  Remarks,  Art.  19.) 

By  exactly  the  same  method  as  in  Art.  34,  it  may  be 
shown  that 

1+»+Ii+a+-"+5i)' 

1  +  -  J    for  positive  integral  values  of  n. 

It  can  be  shown  that  the  limit  of  this  series  is  a  finite 
number  for  all  finite  values  of  x  no  matter  how  great  n  may 
be.     (See  Art.  213  and  Ex.  5,  Art.  215.) 

Corollary.  —  Limf )  =  1,  which  may  be  put  in  the 

form,  lim  ( — ; ]  =  1. 

37.   Derivation  of  [X].  — 

Let  z  =  yx,  then  loge  z  =  x  loge  y.  (y  positive  and  inde- 
pendent of  x.) 

d(\ogez)  =  d(x\og€y), 

dz      ,  7     .      dy 

—  =  \ogeydx  +  x-^; 
&  y 

.*.    dz  =  d  (yx)  =  yx loge  y  dx  +  xyx~l  dy  (y  positive).     [X] 

Note.  —  Formulas  [VII],  [IXa],  [IX&]  are  seen  to  result 
from  [X]  as  special  cases. 


54  DIFFERENTIAL  CALCULUS 

Let  y  =  xn,  then  \ogey  =  n\ogex, 

d(\ogey)  =  d(n\ogex), 
dy  _     dx  < 

y         x  ' 

.'.     dy  =  d  (xn)  =  nxn~l  dx.     [VII] 

If  x  were  negative,  to  avoid  logarithms  of  negative  num- 
bers, both  members  of  y  =  xn  are  squared  before  differ- 
entiating. 

This  derivation  of  [VII]  includes  the  case  where  n  is 
incommensurable. 

38.  Modulus.  —  In  Art.  33  (ii),  it  appears  that  log6  e  is 
the  modulus  of  the  system  of  logarithms  whose  base  is  b. 
Honce,  when  the  base  is  10,  as  in  the  common  system,  and 
the  value  of  e  is  known,  a  table  of  logarithms  will  give  the 
value  of  the  modulus  of  the  common  system  to  as  many 
decimals  as  the  table  gives.  The  modulus  of  the  common 
system,  denoted  by  M,  is  logio  e  =  0.43429  ...  If  this 
value  of  M  is  deduced  independently  of  any  knowledge  of 
the  value  of  e,  which  can  be  done;  then  the  value  of  e  can  be 
gotten  from  a  table  of  logarithms;  for,  since  M  =  logio  e, 
then  e  =  10M;  that  is,  e  is  the  number  whose  common 
logarithm  is  0.43429.  .  .  . 

In  Art,  35  (i)  and  (ii),  it  appears  that  loge  b  is  equal  to 

loge  10  =  .— *—  =  —5 =  2.3026 


log*?'  fee  log10e      .434 

approximately.     (See  Ex.  6,  Art.  215.) 

To  get  these  results  independently,  let  x  be  any  number 
whose  logarithm  in  the  system  with  base  10  is  I,  and  in  that 
with  base  e  is  V;  then  10'  =  x  and  e1'  =  x; 

.*.     10'  =  er.  (1) 

Let  W1  =  e;  (2) 

/.     10'=  1QM;     /.     1=  Ml'  or    p  =  M ,  (3) 


MODULUS  55 

and  since  10  and  e  are  constant,  so  also  is  M.  From  (2), 
M  =  logi0e,    or   from  (1)  I  =  logi0er  =  lf\ogwe; 

j,  =  logi0e  =  M;    or  in  general,  log&e  =  ra. 

Since  I  =  Ml',  or  logio  x  =  M  loge  x,  it  follows  that  the  com- 
mon logarithm  of  any  number  is  equal  to  M  times  the 
Napierian  logarithm  of  that  number. 

Now      d  (log10  x)  =  Md  (loge  x)   or    d^wX^  =  M, 

d  (loge  x) 

or  m  for  base  b,  and  dividing  [VII Ia]  by  [VII lb]  gives 
,  n        v  =  m.  the  modulus; 

d  (l0ge  X) 

/.     M  =  logic  ^  =  modulus  of  common  system  (b  =  10), 
and  m  =  loge  e  =  1 ,  modulus  of  natural  system  (b  =  e). 

From  (1)  above,  Z  loge  10  =  V  or  j,  =  ^jk  =  M,  by  (3) ; 

••    logl° e  =  io^To'  or  loge  10  =  Fi  =  oaMTT.  =  2'3026' 

approximately. 

To  summarize  in  two  equations: 

Common  log  =  0.434  times  natural  log. 
Natural  log     =  2.3026  times  common  log. 

Note.  —  Since  the  modulus  of  the  natural  system  is  unity 
the  differentials  of  logarithms  are  simpler  when  the  logarithms 
are  in  that  svstem;  hence,  in  the  Calculus  and  in  most 
analytic  work,  Napierian  logarithms  are  employed  for  the 
most  part.  Any  finite  number  except  one  could  be  made 
the  base  of  a  system  of  logarithms.  For  computation  the 
common  logarithms  are  the  best,  as  having  the  base  10 
affords  rules  for  the  integral  part  of  the  logarithms  and 
obviates  the  necessity  of  that  part  appearing  in  the  tables. 
It  is  usual  in  writing  log  for  logarithm  to  omit  the  subscript 


56  DIFFERENTIAL  CALCULUS 

indicating  the  base,  when  no  ambiguity  results.  Hereafter, 
when  no  subscript  to  log  appears,  e  will  be  understood. 

39.  Logarithmic  Differentiation.  —  Exponential  func- 
tions and  also  those  involving  products  and  quotients  are 
often  more  easily  differentiated  by  first  taking  logarithms. 
This  method  which  is  used  in  the  last  two  derivations  (Art. 
35  and  Art.  37)  is  called  logarithmic  differentiation. 

To  derive  [Va],  let  z  =  uy,  then  log  z  =  log  u  +  log  y, 

d(\ogz)  =  f  =  ^  +  ?  =  d(logu)  +d(\ogy); 
z         u,         y 

:.     dz  =  d  (uy)  =  ydu  +  udy. 

To  derive  [V&],  let 

V  =  uyz,   then  log  V  =  log  u  +  log  y  +  log  z, 

j/i      t/\      dV      du      dy      dz 
d(logV)  =  T=-  +  7  +  7; 

.'.     dV  =  d  (yz)  =  yzdu  +  uz  dy  -f-  uy  dz. 
To  derive  [VI],  let  z  =  y/x,  then  log  z  =  log  y  —  log  xf 

d  (log  z)  =  -^  =  -;-v  =  f/  (loS2/)  -  d  (log  a;); 

Z  //  O/ 


h-*©- 


|m  =  ^_  ydx  =  xdy-ydx 

x         x2  x2 


40.  Relative  Rate.  Percentage  Rate.  —  The  logarithmic 
derivative  of  a  function  may  be  defined  as  the  relative  rate 
of  increase  of  the  function.     Thus,  when  y  =  f  (x), 

dy 

—  =  •      x    is  the  relative  rate  of  y. 
V       I  {x) 

Hence,   when  z  =  xy  and  therefore,  log  z  =  log  x  -f  logy; 

dz      dx      dy 

dx  _  dx      dx. 

z  "  x        y  ' 


EXERCISE  57 

that  is,  the  relative  rate  of  increase  of  a  product  is  the  sum  oj 
the  relative  rates  of  increase  of  the  factors.  If  the  logarithmic 
derivative  is  multiplied  100  times,  the  product  expresses  the 
percentage  rate  of  increase.     Thus  when 

dy 

^■=Ky,      100-  =  100  K 
ax  y 

is  the  percentage  rate  of  increase,  and  is  here  constant. 


EXERCISE   IV. 

By  one  or  more  of  the  formulas  [I]  to  [X],  differentiate: 

i  ,         ,       ,,  dy      3  log6  e      3  ra 

1.  y  =  log!,  x3  =  3  logft  x.    ,    -j-  —  — - —  =  —  • 

(tx  x  x 

n  n  \s  dy      Slogioe,,  ,,      1.302... 

2.  y  =  (logw  x)\  -£  =  — ^—  (logio  xY  = 


fx  =  -f^  (logio*)2  =     •     x"-  (logic*)'. 

|=log,  +  l. 
dy  _       1  +  log  z 


3.   y  =  s  log  x.  ~-  =  log  a;  +  1. 


4.   w  : 

zlogz  dx  (zlogz)2 

6- » =  1o«  srl  ■ log  {ax  - 6)  - log  (a* + b)-  %~  5CT 

6.   y  =  log  1  +      _  =  log  (1  +  Vx)  -  log  (1  -  Vx). 

8.  7/  =6*e*  &-  (1+  log  6)fc*e*. 

o  i      /  i   i   irx  dy      ax  log  a  +  6*  log  6 

9.  y  =  log  (a«  +  IF).  J!  =  ga.  +  6x  • 

10.  y  =  x>5x.  ^  =  z*54(5+*log5). 

11.  y=a*  ^  =  x*Qogz  +  l). 

12.  y  =  ^.  #=xevl±l^. 

ax  x 

*  In  some  of  the  examples  logarithmic  differentiation  is  employed  to 
advantage;  that  is,  take  logarithms  first  and  then  differentiate. 


58  DIFFERENTIAL  CALCULUS 

Here    log  y  =  ex  log  x     :.      ^  =  ex  —  +  log  xex  dx. 

y         x 

13.  y  =  x]°£  x.  dy  =  2  xlog  «-»  log  x  •  dx. 

14.  y  =  log  (logs).  -£  =  — 

dx      x  log  x 

15.  z,oga  =  alogz.     (Differentiate^both  members  and  verify  results.) 

16.  (x  +  exY  =2^  +  4  xHx  +  6 x2e2X+  4  ze3X  +  e4X.  (Do  as  in  15.) 

17.  (ex  +  e~xy  =  e2X  +  2  +  e~*x.     (Do  as  in  15.) 
ex  —  e~x  dv  4 


18. 


19.   y  =  log 


ex  +  e~x  dx       (ex  +  «-x)2 

ex  dy  1 


1  +  ex  dx      l+ex 


20.  y  =  (log  x)x.  d£  =  (log  x)x  [^-x  +  log  log  x}  • 

21.  Find  the  slope  of  the  curve  y  =  logio  x,  or  x  =  lO2',  showing  that 
the  results  are  identical.  What  is  the  value  of  the  slope  at  (1,  0)? 
What  is  the  slope  of  the  curve  y  =  loge  x,  or  x  =  ey,  and  its  value  at 
(1,0)? 

22.  Find  the  slope  of  the  curve  x  =  loge  y,  or  y  =  ex;  and  note  that 
the  slope  at  any  point  has  the  value  of  the  ordinate  at  that  point.  Value 
of  the  slope  atz  =  0?     At  x  =  1?     Atz=-oo? 

23.  Find  the  slope  of  the  curve  y  =  -  \e "  +  e   <*/  at  x  =  0. 

Ans.  0. 
What  is  the  abscissa  of  the  point  where  the  curve  is  inclined  45°  to 
the  ar-axis  ? 

Ans.  x  =  a  log*  (l  +  V2). 
24.*   Find  the  value  of  x  when  logio  x  increases  at  the  same  rate  as  x. 

Ans.  x  =  log10  e  =  0.4343  .  .  . 

dx 
*  Since  d  (logio  x)  =  logio  e  •  —  ; 

x 

dx=x>  d.(1°glo3:)  =  2.3026 x  •  d  (logio x) ; 
logio  e 

hence,  any  number  N  increases  about  2.3  AT  times  as  fast  as  logio  N. 
When 

N  =  0.4343  .  .  .  ,dN  =  0.4343  X  2.3026  d  (logio  N)  =  d  (logio  A). 

Find  how  much  faster  x  is  increasing  than  logio  x  for  x  =  1. 

Ans.  2.3026  x  =  2.3026  .  .  . 


RELATIVE  ERROR  59 

25.  When  the  space  passed  over  by  a  moving  point  is  given  by 
s  =  ael  +  be*1,  find  the  velocity  and  the  acceleration,  showing  that  the 
acceleration  is  equal  to  the  space. 

26.  Find  the  slope  of  the  curve  y  =  -\e"  —  e    «/  at  x  =  0. 

Ans.  1. 

X  _x 

ga  g    a  \ 

27.  Find  the  slope  of  the  curve  y  = at  x  =  0.     Ans.  -• 

ea  +  e   a 

28.  Find  the  derivative  of  the  implicit  function  y  in  (?+v  =  xy. 
Passing  to  logarithms : 

i  i  ,  dy      y  (1  -  x) 


29.  xy  =  yx.     Passing  to  logarithms : 

y  log  x  =  x  log  y.         ^| 

30.  Find  the  slope  of  the  probability  curve  y  =  e~*2. 


.«~-»>™     t-&g& 


Ans.   -2xe-x\ 

What  is  the  value  of  the  slope  at  z  =0? 

Ans.  0. 

2 

At  I  =  1?  Ans. 

e 

41.   Relative  Error.  —  Since  when  y  =  f(x),  the  relative 
rate  of  increase  of  y  is 

dy      df(x)  . 

dx  _     dx     _  f  (x) 

where  dy  =  /'  (x)  dx ; 

hence,  Ay  =  f  (x)  kx,  (1) 

are  approximate  relations.     The  relation   (1)   is  useful  in 

finding  the  error  in  the  result  of  a  computation  due  to  a  small 

error  in  the  observed  data  upon  which  the  computation  is 

based.     The  relation   (2)   gives  approximately  the  relative 

Aty 
error  — -• 

y 


60  DIFFERENTIAL  CALCULUS 

1.  Thus,  to  find  an  expression  for  the  relative  error  in  the 
volume  of  a  sphere  calculated  from  a  measurement  of  the 
diameter  when  there  is  an  error  in  the  measurement.     Here 

.      A  7      (I        \   AD  AD 

Hence,  an  error  of  one  per  cent  in  the  measurement  of  the 
diameter  gives  approximately  an  error  of  three  per  cent  in 
the  calculated  volume. 

2.  Again,  from  the  formula  for  kinetic  energy  K  =  \  mv2, 
to  show  that  a  small  change  in  v  involves  approximately 
twice  as  great  a  relative  change  in  K.     Here 

AK  v  Av       nAv 

K  §  mv2  v 

3.  If  a  square  is  laid  out  100  ft.  on  a  side  and  the  tape  is 
0.01  ft.  too  long,  an  error  of  TJ^y  of  one  per  cent,  the  relative 
error  in  the  area  is,  approximately, 

M_  ox^  -200-^1-  0002 
A    ~2X  x2  -2UO(100)2"'UUU2' 

or  Ti o  of  one  per  cent. 

42.   The  Compound  Interest  Law.  — The  limit  (l+-j 

=  ex,  in  Art.  36,  arises  in  a  variety  of  problems.  When  a 
function  has  the  general  form 

y  =  aebz,  (1) 

then  -^  =  baebx  =  by; 

that  is,  the  rate  of  ch^ge*  of  the  function  is  proportional  to 
the  function  ifself.  Many  of  the  changes  that  occur  in 
nature  are  in  accordance  with  this  law,  called  by  Lord  Kelvin 
the  compound  interest  law.  It  is  so  called  because  of  the  fact 
that  the  amount  of  a  sum  of  money  at  compound  interest 
has  a  rate  of  change  at  any  value  proportional  to  that  value, 
when  the  change  is  continuous. 


THE  COMPOUND   INTEREST  LAW  61 

Let  A  =  amount,  r  =  rate  per  cent,  P  =  principal,  and  t 
number  of  years;   then, 

A  =  P(l  +  r)1,  when  interest  is  compounded  yearly; 
at  n  equal  intervals  each  year; 


K" 


A  =  lim  P 


(l+^nJ  =  Pe'<(byArt.  36)  (2) 


is  the  amount  when  the  interest  is  compounded  continuously. 

dA 

Here  A  =  Pert  and  -4r  =  rPert  =  rA,  hence,   the   rate   of 
dt 

change  of  the  A  is  proportional  to  the  value  of  A ,  the  factor 

of  proportionality  being  the  rate  per  cent  at  which  the 

interest  is  reckoned.     As  a  comparison,  it  may  be  noted  that 

$1.00  will  amount  to  $2,594,  in  ten  years  with  interest  at  10 

per  cent,   compounded  yearly;    while  the  amount  will  be 

$2,718  when  compounded  continuously. 

If  in  A  =  Pert,  t  increase  in  any  arithmetical  progression, 
whose  common  difference  is  h,  A  will  increase  in  a  geometrical 
progression  whose  common  ratio  is  erh;  for  if  t  become  t  +  h, 
A  will  become  Per(t+h\  that  is,  Aerh.  Hence  A  is  a  quantity 
which  is  equally  multiplied  in  equal  times. 

The  density  of  the  air  towards  the  sea  level  from  an  eleva- 
tion is  a  quantity  which  is  equally  multiplied  in  equal 
distances  of  descent,  for  the  increase  in  density  per  foot  of 
descent  is  due  to  the  weight  of  that  layer  of  air  which  is  itself 
proportional  to  the  density.  Manys  other  instances  occur 
in  physics.  -~f 

When  bacteria  grow  freely  the  increase  per  second  in  the 
number  in  a  cubic  inch  of  culture  is  proportional  to  the 
number  present.  The  relation  between  the  number  N  and 
the  time  t  is  expressed  by  the  equation, 

N  =  Cekt>   •'•    w  =  kCekt  =  kNi  (3) 


62  DIFFERENTIAL  CALCULUS 

where  N  is  the  number  of  thousand  per  cubic  inch,  and  k 
is  the  rate  of  increase  shown  by  a  colony  of  one  thousand  per 
cubic  inch.  So  many  instances  of  this  kind  are  found  in 
organic  growth  —  where  the  rate  of  growth  grows  as  the 
total  grows  —  that  the  law  is  called  the  law  of  organic  growth, 
as  well  as  the  compound  interest  law. 

When  a  quantity  has  a  rate  of  change  which  is  proportional 
to  the  quantity  itself,  if  the  functional  relation  is  expressed 
by  an  equation,  it  must  be  of  the  form  (1). 

In  the  case  of  the  density  of  the  air,  the  relation  (see  Art. 
226)  between  the  density  p  and  the  height  h  above  the  sea 
level  is  expressed  by 

P  =  po6~kh>      '"'      a%=  ~  kP»e~kh=  ~  kP>  W 

where  p0  is  the  density  at  the  sea  level  and  k  is  a  constant  to 
be  determined  by  experiment.  From  barometric  observa- 
tions at  different  altitudes,  it  has  been  found  that  at  the 
height  of  3J  miles  above  the  earth's  surface,  the  air  is  about 
one-half  as  dense  as  it  is  at  the  surface.  Hence,  to  deter- 
mine k, 


p-s.bk  —  I  ; 

6            25 

-3.5  k 

=  log  0.5, 

or     k  = 

log  0.5 
-3.5 

0.198; 

•*•  P  = 

Poe-°-198\ 

where  h  is 

>  in  miles. 

(5) 

Here,  as  h  increases  in  arithmetical  progression,  p  decreases 
in  geometrical  progression,  the  force  of  gravity  and  the 
temperature  being  taken  constant.  The  varying  density  at 
different  heights  is  found  by  giving  values  to  h ;  thus,  making 

h  =  35,  gives  —  =0.001;  hence,  according  to  this  law  at  the 
Po 

height  of  35  miles  the  density  of  the  air  is  about  one-thou- 
sandth of  the  density  at  the  sea  level.     As  the  pressure  p  is 


THE  COMPOUND   INTEREST  LAW  63 

proportional  to  the  density,  p  =  k'p;  and  the  same  law  holds 
for  the  pressure  of  the  air;  hence, 

p  =  poe-k'h,     :.      -|  =  -kpoe-k'h  =  -k%  (6) 

where  kf  is  a  constant  to  be  found  by  experiment. 

Knowing  the  pressure  at  the  sea  level  and  observing  the 
pressure  at  some  height,  kf  is  determined;  or  it  can  be 
determined  from  the  value  of  the  pressure  at  any  two  differing 
heights.  When  the  pressure  is  expressed  in  inches  of  mercury 
in  a  barometer,  the  pressure  in  lbs.  per  square  inch  =  0.4908 
X  barometer  reading  in  inches.  Taking  p0  =  30"  when 
h  =  0,  and  p  =  24",  say,  when  h  =  5830  ft.,  k'  is  readily 
computed.  In  millimeters  the  equation  is  p  =  760  e~h/mo, 
where  h  is  in  meters. 

The  relation  between  the  decomposition  of  radium  and 
time  is  expressed  by  the  equation 

q  =  qtfr";     .'.      J  =  -kq^~kt  =  -  kq,  (7) 

where  q0  is  the  original  quantity  and  q  is  the  quantity  remain- 
ing after  a  time  t.  The  constant  k  can  be  found  from  the 
fact  that  half  the  original  quantity  disappears  in  1800  years. 
The  relation  between  the  varying  difference  of  tempera- 
ture of  a  body  and  that  of  the  surrounding  medium  and  the 
time  of  cooling  is  expressed,  according  to  Newton's  Law,  by 

b  =  8oe-kt;     .'.      ~  =  -k80e~kt  =  -kd,  (8) 

where  8  =  r  —  r0,  the  difference  in  the  temperature  of  the 
body  and  that  of  the  medium,  50  =  t\  —  r0,  the  difference 
when  t  =  0,  k  a  constant;  that  is,  r  =  r0  +  (ti  —  r0)  e~kt, 
where  —  kt  indicates  the  body  is  cooling. 


64  DIFFERENTIAL  CALCULUS 

TRIGONOMETRIC   FUNCTIONS. 

43.  Circular  or  Radian  Measure.  —  The  formulas  for 
differentiation  of  trigonometric  functions  are  simpler  when 
the  angle  is  measured  in  radians  than  in  degrees.  Hence, 
in  the  formulas  that  follow,  the  angle  will  be  in  radians. 

A  radian,  the  unit  of  circular  measure,  is  an  angle  which 
when  placed  at  the  center  of  a  circle  intercepts  an  arc  equal 

in  length  to  the  radius. 

180° 
Since  2  wr  is  arc  of  360°,  a  radian  equals  ,  or  57.3° 

♦  7T 

nearly.  In  circular  or  radian  measure,  an  angle  in  radians 
is  equal  to  the  length  of  the  intercepted  arc  divided  by  the 

radius;    0  =  -,  where  6  is  angle  in  radians,  s  is  number  of 

units  in  arc,  and  r  is  the  number  of  units  in  radius.  Hence, 
s  =  rd;  that  is,  in  any  circle  the  length  of  an  arc  equals  the 
product  of  the  measure  of  its  subtended  central  angle  in 
radians  and  the  length  of  the  radius.  If  r  =  1,  then  s  =  d; 
that  is,  the  arc  and  the  angle  have  the  same  numerical 
measure.  Trigonometric  functions  are  called  circular  func- 
tions. 

44.  Formulas  and  Rules  for  Differentiation.  — 

[XI]  d*  sin  8  =  cos  6  d$. 

The  differential  of  the  sine  of  an  angle  is  the  cosine  of  the 
angle  by  the  differential  of  the  angle. 

[XII]  dcos6  =  -  sin0c/e. 

The  differential  of  the  cosine  of  an  angle  is  minus  the  sine  of 
the  angle  by  the  differential  of  the  angle. 

[XIII]  dtanG  =  sec26d9. 

The  differential  of  the  tangent  of  an  angle  is  the  secant 
squared  of  the  angle  by  the  differential  of  the  angle. 

[XIV]  d  cot  9  =  -  cosec2  9  </8. 

*  Parenthesis  after  d  omitted  when  no  ambiguity  results. 


DERIVATION  OF   [XI]  AND   [XII] 


65 


The  differential  of  the  cotangent  of  an  angle  is  minus  the 
cosecant  squared  of  the  angle  by  the  differential  of  the  angle. 

[XV]  d  sec  6  =  sec  9  tan  6  d9. 

The  differential  of  the  secant  of  an  angle  is  the  secant  of  the 
angle  by  the  tangent  of  the  angle  by  the  differential  of  the  angle. 

[XVI]  d  cosec  0  =  -  cosec  6  cot  9  d9. 

The  differential  of  the  cosecant  of  an  angle  is  minus  the  cose- 
cant of  the  angle  by  the  cotangent  of  the  angle  by  the  differential 
of  the  angle. 

45.   Derivation  of  [XI]  and  [XII].  — 

I.  Let  the  point  P(x,  y)  move  along  the  arc  XPY  of  a  unit 
circle.  Denote  the  number  of  linear  units  in  the  arc  XP  by 
s,  and  the  number  of  radians  in  angle  XOP  by  0. 

Then  0  =  s,        y  =  sin0,         x  =  cos  0; 

:.     dd  =  ds,     dy  =  d  sin  0,     dx  =  d  cos  0. 
Angle  DTP  equals  6,  and  dx  is  nega- 
tive;   hence,  from  the  triangle  DTP, 
by  (d)  Art,  10, 

dy  =  d  sin  d  =  cos  Odd,     since  ds  =  dd; 

dx  =  d  cos  0  =  —  sin  6  dd, 

since     —dx  =  sin  6  ds  and  ds  =  dd. 

It  is  seen  that  dy  and  dx  in  the  figure 
are  what  the  changes  of  the  sine  and 
cosine  of  6  would  be  if,  at  the  value  XOP  of  6,  the  changes 
were  to  become  uniform. 

II.  By  the  general  method  of  limits. 

Let       y  =  sin  d,    then    y  +  Ay  =  sin  (d  +  A0) ; 


by  =  sin  (0  +  A0) 
A0 

eir> 

by 


sin  0  =  2  sin 


sin 


A0 


A0 
2 


cos 


(•+¥> 


Trig.; 


[(Art.  46). 
=  1,  as  A0  =  0; 


66 


DIFFERENTIAL  CALCULUS 

A0^ 


sin 


-7T      =  COS  i 


d0       a^oL^J      a^o      A?     I        V         2/ 

,\     cfy  =  d  sin  0  =  cos  0  dd. 
Corollary.  —  d  covers  0  =  d  (1  —  sin  0)  =  —  cos  0  dd. 

Now  let         z  =  cos  0  =  sin  (~  —  0j ; 

d#  =  dcos0  =  dsinl^  —  0j  =  cosf^  —  djdl-  —  6j; 

'.     d  cos  0  =  —  sin  0  d0. 
Corollary. —  d  vers  0  =  d  (1  —  cos  0)  =  sin  0  c?0. 

LJj  (=!)  =  , 

Let  0  be  the  number  of  radians  in  the  angle  NO  A,  where 
the  angle  is  taken  acute;   by  Ge- 
ometry, if  AT  and  BT  are  tan- 
gents at  A  and  B, 
then, 

chord  AB  <  arc  AB  <  AT  +  BT, 
and  therefore 

MA  <  arc  iVA  <AT. 
Hence 


MA      arciVA      AT 
OA  <     OA      <  OA  ' 


that  is, 

dividing  by  sin  0, 


sin0  <0  <  tan0; 


K      7i<JL2    or   1>-"->cos0. 

sin  0      cos  0  0 


Thus  the  ratio 


0 
in 

sin  0 


0 


lies  between  1  and  cos  0. 


LIMIT  OF  RATIO  OF   INFINITESIMALS  67 

When  0  approaches  0  as  its  limit,  cos  0  approaches  1  as  its 
limit;  therefore,  also  sin  0/0  approaches  1  as  its  limit.  (See 
Art.  215,  Ex.  1.) 

Corollary.  —  Since     lim  — —  =  1, 

0=0     0 

..      2  MA      ..      chord  AB      t      ,.       M    ,     , 

hm      A^      =  hm  j^-  =  1.     (Art.  22,  also.) 

It  may  be  noted  that, 

MA      sin  0  „,,..«, 

since     -ryF  =  t — «  =  cos  0,     when   0  =  0  and  cos  0=1, 
A T      tan  0 

sin0  and  tan  0  approach  equality;  and,  since  the  arc  0  is 

intermediate  in  value  between  sin  0  and  tan  0,  the  three 

functions  approach  equality  as  the  angle  0  nears  zero.     So 


lim  ( - — - )  =  1  and  lim  [— — )  =  1,  as  well  as  hm 
6=o  \tan0/  0=0  V    0    /  0=0 


/sin 
11  V~T 


=  1. 


These  are  fundamental  examples  of  the  ratio  of  infini- 
tesimals approaching  a  constant  value  as  a  limit.  Con- 
sider again  the  equality  of  ratios,  -t-=-  =  -^nr'     Suppose  the 

points  A  and  B  approach  N;  so  long  -as  A  and  B  are  not 
coincident,  that  is,  so  long  as  AB  is  really  a  chord,  the 
equality  still  exists.  The  ratio  MA  :  A  T  may  be  considered 
a  function  of  OM,  or  equally  well  a  function  of  the  angle 
NO  A.  As  03/  approaches  ON  as  its  hrnit,  or  as  the  angle 
NO  A  approaches  zero  as  a  limit,  the  ratio  MA  :  AT  ap- 
proaches 1  as  its  limit.  The  nearer  OM  gets  to  ON,  or  the 
nearer  A  gets  to  N,  the  nearer  does  the  ratio  MA  :  A77  get 
to  unity.  The  crucial  fact  is  that  the  reasoning  is  vitiated 
if  OM  becomes  actually  equal  to  ON;  for  then  the  triangles 
will  cease  to  exist,  the  terms  of  the  one  ratio  will  be  zero  and 
those  of  the  other  will  be  identical,  and  the  equation  on 
which  the  reasoning  is  based  could  not  be  established. 


68  DIFFERENTIAL  CALCULUS 

47.   Derivation  of  [XIII].  — 

Since  tan0  = -,     dtsaid  =  d( -J; 

Vcos0/' 


d  tan  0  = 


COS0' 

cos  0  d  sin  0  —  sin  0  d  cos  0 

cos20 
(cos20  +  sin20)d0 


cos20 
48.   Derivation  of  [XIV].  - 
Since       cot0  =  tan 


=  sec2  0  dd. 


%->)■■ 

/.     dcot0  =  dtan(|  -  0)  =  sec2  (|  -  fl)d  fe  -  d) 


=  —  cosec2  0d6 
49.   Derivation  of  [XV].  — 
Since  sec0  = 

COS0 


sin  0  c?0 

d  sec  0  =  d = — 

cos20 

=  sec  0  tan  0  d0. 


\cos  0/ 

=  £ 

50.   Derivation  of  [XVI].  - 

Since  cosec  0  =  sec  (^  —  0] ; 

/.     dcosecfl  =  dscc(|  -  0)  =  sec/|  -  0)  tan  fe  -  d)  d  fe  -  d) 
=  —  cosec  0  cot  0  dd. 

Note.  —  In  the  derivations  of  the  formulas  for  the  cosine, 
cotangent,  and  cosecant,  as  given,  it  may  be  noted  that,  as 
in  the  last  example  of  Art.  31,  the  formula  for  the  derivative 
of  the  function  of  a  function  has  appropriate  application. 

Thus  for  cos0,  let  x  =  cos0  =  sin(~  —  0J  and  0  =  ^  —  0, 


NOTE  ON   [XI]  69 

dx      dx    d<f> 


then, 


or  Is 


.'.     -rn  cos  6  =  —  sin  6    or    d  cos  0  =  —  sin  B  dQ* 
dd 


In  practice  the  actual  substitution  of  the  auxiliary  symbol  <j> 
may  be  dispensed  with. 

The  formula  applies  to  such  functions  as  y  =  sin  (ax  +  b) . 
Thus  put  z  =  ax  +  b,  making  y  =  sin  z;  then 

dy  _  dy    dz 
dx      dz    dx 

or        -=-  sin  (ax  +  6)  =  -=- sin  2  X  -=-  (ax  -f-  6)  =  cos  2  •  a: 

.*.     -=-  sin  (ax  +  6)  =  a  cos  (ax  +  5) 
or  d  sin  (ax  +  b)  =  a  cos  (ax  +  6)  dx. 

Again,  let  the  function  be  y  =  sin2  (ax  +  0)  and  put  z  =» 
sin  (ax  +  b) ;  then 

^  sin2  (ax  +  6)  =  —  (z2)  .  —  sin  (ax  +  6) 

=  2  2  X  a  cos  (ax  +  6) 
=  2  a  sin  (ax  +  6)  cos  (ax  +  6) ; 
.'.     d  sin2  (ax  +  6)  =  2  a  sin  (ax  +  6)  cos  (ax  +  b)  dx. 

51.   Note  on  [XI].  —  If  the  angle  is  measured  in  degrees, 

then  d  sin  6  =  —  cos  6  d0,  since  0  degrees  is  r-^r  radians 

lot)  loU 

and  sin0°  =  sin^^j; 


d  sin  0C 


=  dsin&) 

7T  /7T0r\    -„  7T  „0    ,„ 

=  T80COS(lWcW=180COS<'d(>- 


70  DIFFERENTIAL  CALCULUS 

It  is  thus  seen  that  the  formulas  for  differentiation  of  the 
trigonometric  functions  are  simpler  when  the  angle  is 
measured  in  radians  than  when  measured  in  degrees.  For 
the  same  reason  that  Napierian  or  natural  logarithms  are 
employed  in  differentiation,  radian  or  circular  measure  is 
used  for  angles  of  the  trigonometric  functions,  when  differ- 
entiation is  to  be  done. 

52.   Remarks  on  [XI].  —  The  fundamental  limit  of  Art. 

46,  lim(-T—  )=  1  means  that  when  0  is  a  small  number 
0=0  V   0   /       

sin  0  is  approximately  equal  to  0.     For  angle  of  1°,  0  =  j^: 

=  0.0174533  .  .  .  ,  sin  0  =  0.0174524  .  .  .  ;  so  "tney  are 
equal  to  five  decimals.  Of  course  for  angle  of  1'  or  1",  they 
are  equal  to  a  great  many  more  decimals,  but  theyaTe  never 
exactly  equal  however  small  the  angle  may  be,  since  the  sine 
is  always  less  than  the  arc. 

Since  d  sin  0  =  cos  0  dd,  if  the  value  0°  is  taken  for  0  and 


dsind  =  cos0°~  =  .0174533  ...     or     ^^  =  1, 

A  sin  0  =  sin  (0°  +  A0)  -  sin  0°  -  .0174524  .  .  .   =  sin  1°; 

A  sin B  _  .0174524  .  .  .  ..     A  sinfl  =  rising  _    1 

A0     "  .0174533  .  .  .        '    US     A0  dd 

6=0° 

From       ,      =  cos0=  cos0°  =  1,  it  is  seen  that,  at  0  =  0°, 
do 

the  sine  of  0  is  changing  at  the  same  rate  as  0  is  changing;  so 

the  slope  of  the  curve  y  =  sin  0  is  unity  at  the  origin,  and  the 

tangent  to  the  curve  at  that  point  makes  an  angle  of  45°  with 

0-axis.     The  conditions  are  the  same  at  0  =  2  -k.     As     " 

do 

IT 

=  cos0  =  cos 90°  =  0,  at  0  =  ^,  the  sine  of  0  is  not  changing, 


REMARKS  OX    [XI] 


71 


the  rate  being  zero,  and  the  tangent  to  the  curve  at  that 
point  is  parallel  to  0-axis.  At  6  =  it,  cos  180°  =  —  1,  so  the 
sine  of  0  is  decreasing,  at  that  value  of  0,  at  the  same  rate  as 
0  is  increasing,  and  the  tangent  to  the  curve  at  that  point 
makes  angle  of  135°  with  0-axis.  Thus  the  rate  of  change  of 
the  sine  of  0,  at  any  value  of  0,  can  be  found ;  and  the  difTeren- 


sin  6 


tial  of  the  sine,  the  change  if  the  change  became  uniform, 
will  always  differ  from  the  increment,  the  actual  change  of 
the  sine,  when  the  angle  is  given  an  increment.  In  taking 
sines  or  other  functions  from  tables  by  interpolation,  the 
changes  are  assumed  as  uniform  within  allowable  limits  of 
error. 


EXERCISE  V. 


1.  y  =  sin  x-. 

2.  y  =  sin2  x. 

3.  y  =  cos  ax. 

4.  y  =  /(0)  =  tanw0. 

6.  f{6)  =  tan30  +sec30. 

6.  /  (as)  =  sin  (log  ax) . 

7.  f(x)  =  log  (sin  ax). 

Q  sin  x  +  cos  x 

8.  y= 

9.  /(0)  =  log  (tan  ad). 

10.  f(d)  =  log  (cot  00). 

11.  /(0)  =  tan  (log  0). 

12.  f(d)  =  log  (sec  0). 


dy  =  2  x  cos  x2  dx. 

dy  =  2  sin  x  cos  x  dx  =  sin  2  x  dx. 

dy  =  —  a  sin  ax  dx. 

-M  =f  (0)  =m  tan™"1 0  sec2  0. 


fix) 
fix) 
dy  = 
dx 

r® 

f(e) 

f'(0) 

=  3  sec2  3  0  +  3  sec  3  0  tan  3  0. 
=  1/xcos  (log ax). 
=  a  cot  ax. 
2sinz 

2a                       2a 

~  2  sin  (off)  cos  (ad)  ~~  sin  (2  ad) 
=  -  2  a/sin  (2  a0). 
=  l/0sec2(log0). 

/'  (e)  = 


sec  0  tan  i 
sec0 


=  tan0. 


72  DIFFERENTIAL  CALCULUS 

13.  /  Or)  =  xsin  x.  f  (x)  =  xsin  x  (sin  x/x  +  log  x  •  cos  x). 

14.  /  (re)  =  (sin  x)x.  f  (x)  =  (sin  x)x  {x  cot  x  +  log  sin  x). 

15.  /  (0)  =  (sin  0)tan  e.  f  (0)  =  (sin  0)tan  e  (1  +sec2  0  log  sin  0) . 

16.  /(0)  =  |  tan3 0- tan 0+0.  /'  (0)  =  tan40. 


„  ,  ,        -(sec  Vl  -  x)2 
3  {X)            2Vl-x 

r.  f{x)  =«  tan  Vl  -x. 

>•/«»=  log  \/f^- 

/'  (0)  =  esc  0. 

By  differentiation  derive  each  of  the  following  pairs  of  identities 

from  the  other: 

19.   sin  2  0  =  2  sin  0  cos  0,  cos  2  0  s  cos2  0  -  sin2  0. 

on      .    na         2tan0  0.       1  -  tan2 0 

•20.    sin20  =  — — — — ,     cos20  =  r— — — ~ 

1  +  tan2  0  1  +  tan2  0 

21.  sin  3  0  =  3  sin  0  -  4  sin3  0, 
cos  3  0  =  4  cos3  0  —  3  cos  0. 

22.  sin  (m  +  n)  0  s  sin  to0  cos  ri0  +  cos  md  sin  ri0, 
cos  (m  -\-n)6  =  cos  md  cos  n9  —  sin  m0  sin  n0. 

23.  If  0  vary  uniformly,  so  that  360°  is  described  in  w  seconds,  show 
that  the  rates  of  increase  of  sin  0,  when  0  =  0°,  30°,  45°,  60°,  90°,  are 
respectively,  2,  V3,  VS,  1,  0,  per  second.     (See  figure,  Art.  52.) 

53.  The  Sine  Curve  or  Wave  Curve.  —  The  locus  of  the 
equation 

y  =  sinx,  (1) 

where  x  is  an  angle  in  radians,  is  called  the  sine  curve,  from  its 
equation,  or  the  wave  curve,  from  its  shape.  The  maximum 
value  of  y  is  called  the  amplitude,  being  unity  in  (1);  and, 
since  the  curve  is  unchanged  when  x  +  2  it  is  substituted  for 
x,  the  curve  y  =  sin  x  is  a  periodic  curve  with  a  period  equal 
to  2  ir.  (See  figure,  Art.  52,  and  figure,  Art.  73.) 
The  more  general  form  of  the  equation  is 

y  =  a  sin  mx,  (2) 

where  a  is  the  amplitude  and  —  is  the  period,  m  a  constant. 

The  curve  is  called  the  sinusoid  also,  and  is  of  great 
importance,  since  it  is  the  type  form  of  the  fundamental 
waves  of  science;  such  as,  sound  waves,  vibrations  of  rods, 


DAMPED  VIBRATIONS  73 

wires,  plates  and  bridge  members,  tidal  waves  in  the  ocean, 
and  ripples  on  a  water  surface.  The  ordinary  progressive 
waves  of  the  sea  are  not  of  this  shape,  as  they  have  the 
form  of  a  trochoid. 

54.  Damped  Vibrations.  —  When  a  body  vibrates  in  a 
medium  like  a  gas  or  liquid,  the  amplitude  of  the  swings  get 
smaller  and  smaller,  or  the  motion  slowly  (or  rapidly  in  some 
cases)  dies  out.  Thus,  when  a  pendulum  vibrates  in  the  air 
the  rate  of  decay  of  the  amplitude  is  quite  slow;  but  when 
in  oil  the  rate  is  rapid.  The  ratio  between  the  lengths  of  the 
successive  amplitudes  of  vibration  is  called  the  damping 
factor  or  the  modulus  of  decay. 

In  all  such  cases  the  amplitude  of  the  swings  differ  by  a 
constant  amount  or  the  logarithmic  decrement  is  constant. 
Hence  the  amplitude  must  satisfy  an  equation  of  the  form 

A  =  ae~bt,  (1) 

where  A  is  the  amplitude  and  t  the  time.  The  actual 
motion  is  given  by  an  equation  of  the  form 

v 
y  =  ae~bt  sin  cot,     where  a>  =  -  is  a  constant.         (2) 

(See  Art,  73.) 

*  In  plotting  a  curve  whose  equation  is  of  this  form,  say, 

7T 

y  =  e-<xsin-x,  (3) 

much  is  gained  by  the  following  considerations: 

1.  Since  the  numerical  value  of  the  sine  never  exceeds 
unity  the  values  of  y  in  (3)  will  not  exceed  in  numerical  value 
the  value  of  the  first  factor  e~*z.  As  the  extreme  values  of 
sin|7rx  are  +1  and  —l,y  has  the  extreme  values  e~ix  and 
—  e~*x.     Hence,  if  the  curves 

y  ==  g-k*    and     y  =  —e~ix  (4) 

*  This  illustration  is  given  substantially  in  Smith  and  Gales's  New 
Analytic  Geometry. 


74  DIFFERENTIAL  CALCULUS 

are  drawn,  the  locus  of  (3)  will  lie  entirely  between  these 
curves.  They  are  called  boundary  curves,  and  they  are 
plotted  by  three  or  more  points,  the  second  being  symmetri- 
cal to  the  first  with  respect  to  the  z-axis. 


2.  When  sin  f  irx  =  0,  then  in  (3)  y  =  0,  since  the  first 
factor  is  always  finite.  Hence,  the  locus  of  (3)  meets  the 
a:-axis  in  the  same  points  as  the  sine  curve 

y  =  sin  |  irx.  (5) 

3.  The  required  curve  is  tangent  to  the  boundary  curves 
when  the  second  factor,  sin  J  irx,  is  +1  or  —  1;  that  is,  when 
the  ordinates  of  the  curve  (5)  have  a  maximum  or  a  minimum 
value.  The  tangency  is  proven  by  finding  the  derivative  of 
y  in  (3)  and  noting  that,  when  sin  \  irx  is  +1  or  —  1,  it  will  be 
the  same  as  the  derivatives  of  y  in  (4) .  Hence,  the  slopes  of 
the  curves  and  the  ordinates  being  equal  for  the  same  values 
of  x,  the  required  curve  is  tangent  to  one  or  the  other  of  the 
boundary  curves  for  those  values  of  x  that  make  sin  \  irx  = 
+  1  or  —1.     Thus,  differentiating  (3)  and  (4)  gives 

dy  1     ,     .    1        .  v     .  1 

j-  =  —  t  e_i   sin  7T  irx  +  -  e_i  x  cos  s  irx 
dx  4  2  2  2 

=  —  i  e-i  z,  when  sin  \  irx  =  1, 

=  \  e_i  x,  when  sin  \  wx  =  —  1 . 

For  the  sine  curve  (5)  the  period  is  4  and  the  amplitude  is  1. 
This  curve  is  the  broken  line  of  the  figure. 

The  locus  of  (3)  crosses  the  x-axis  at  x  =  0,  ±2,  ±4,  ±6; 


DAMPED  VIBRATIONS  75 

etc.,  and  is  tangent  to  the  boundary  curves  (4)  at  x  =  ±1,  ±3, 
±5,  etc.  The  discussion  having  disclosed  these  facts,  the 
curve  is  readily  sketched,  as  in  the  figure;  that  is,  the  wind- 
ing curve  between  the  boundary  curves  (4). 

A  more  general  form  of  the  equation  of  a  damped  vibration 
is 

y  =  ae~bt  sin  (ut  —  a),  where  a  =  —  is  constant.      (3') 

This  equation  may  be  written  either  (see  Art.  73) 
y  =  e~bt  (A  sin  kt  +  B  cos  kt), 
where  A  and  B  are  constants, 
or  y  =  A  sin  (cot  —  a),  where  A  =  ae~ht.  (3") 

Here  A  is  a  variable  decreasing  amplitude,  whose  relative 
rate  of  decrease  is  —  dA/dx  -s-  A  =  b;  that  is,  the  relative 
rate  of  decrease  of  A  is  constant. 

The  successive  derivatives  from  (3')  are  (by  Art.  68) : 

dy 

-JL  =  ae~ht  [—  b  sin  (ut  —  a)  +  co  cos  (at  —  a)], 

d2y 

—jL  =  ae-bt  [52^2  sm  (wj  _  a)  _  2  oo)  cos  (ut  —  a)], 

whence  it  follows  that 

Equations  which  contain  derivatives  or  differentials  are 
called  differential  equations.  The  equation  (4')  is  the  funda- 
mental differential  equation  for  damped  vibrations.     The 

term  in  -r-,  or  v,  proportional  to  the  velocity,  occurs  in  equa- 
tions for  vibration  only  when  damping  is  considered.  Vibra- 
tions are  cases  of  simple  harmonic  motion  —  damping  being 
caused  by  resistances,  such  as  friction,  etc.  Simple  har- 
monic motion  is  treated  in  Art.  73. 


76  DIFFERENTIAL  CALCULUS 

INVERSE  TRIGONOMETRIC  FUNCTIONS. 

55.  Formulas    and    Rules    for    Differentiation.  —  The 

direct  trigonometric  functions  are  single-valued  but  the 
angle  has  to  be  restricted  to  a  certain  range  in  order  that 
the  inverse  functions  may  be  single  valued.  To  make 
the  inverse  functions  single-valued,  the  angle  denoted  by- 
sin-1  x,  cosec-1  x,  tan-1  x,  cot-1  x,  covers-1  x,  is  taken  to  lie 

between  —  -  and  ~ ,  and  the  angle  denoted  by  cos-1  x,  sec-1  xy 
vers-1  x}  to  lie  between  0  and  t.    Thus 

^=5=oos_1("2?);  sinr"1  (—  I)  — I- 


V       2  /        6  ' 


sin-1  x  +  cos-1  x  =  ■= ; 
tan-1  x  +  cot-1  x  =  =  ,  if  x  be  positive, 

=  -r,  if  x  be  negative. 

These  restrictions  will  be  assumed  in  the  following  formulas, 
and  all  will  be  expressed  in  terms  of  the  letter  x.  While  the 
symbols  sin-1  x  and  arc  sin  x  are  both  used  to  denote  the 
angle  whose  sine  is  x,  in  writing  the  formulas  the  notation 
sin-1  x  is  preferable. 

[XVII]  d  sin-1  x  =       dx      ■ 

Vl-  x2 

The  differential  of  an  angle  in  terms  of  its  sine  is  the  differen- 
tial of  the  sine  divided  by  the  square  root  of  one  minus  the  square 
of  the  sine. 

[XVIII]  dcos-1*-        — — 


V1-.T2 

The  differential  of  an  angle  in  terms  of  its  cosine  is  minus 
the  differential  of  the  angle  in  terms  of  its  sine. 

[XIX]  </tan-1^  =  T4:L-2- 

1  +  .Y*5 


DERIVATION  OF   [XVII]  AND   [XVIII]  77 

The  differential  of  an  angle  in  terms  of  its  tangent  is  the 
differential  of  the  tangent  divided  by  one  plus  the  square  of  the 
tangent. 

dx 

[XX]  dcotr1*-  -TT^' 

1  +  x2 

The  differential  of  an  angle  in  terms  of  its  cotangent  is  minus 
the  differential  of  the  angle  in  terms  of  its  tangent. 

[XXI]  d  sec"1  x  =  — =       . 

xVx2  -1 

The  differential  of  an  angle  in  terms  of  its  secant  is  the  differ- 
ential of  the  secant  divided  by  the  secant  and  the  square  root  of 
the  square  of  the  secant  minus  one. 

[XXII]  d  cosec"1  x  = dx       . 

The  differential  of  an  angle  in  terms  of  its  cosecant  is  minus 
the  differential  of  the  angle  in  terms  of  its  secant. 

dx 


[XXIII]  rivers"1*  = 

V2x-x2 

The  differential  of  an  angle  in  terms  of  its  versine  is  the 
differential  of  the  versine  divided  by  the  square  root  of  twice  the 
versine  minus  the  square  of  the  versine. 

dv 

[XXIV]  d  covers"1  x  = ax 

V2x-x2 

The  differential  of  an  angle  in  terms  of  its  coversine  is  minus 
the  differential  of  the  angle  in  terms  of  its  versine. 

56.  Derivation  of  [XVII]  and  [XVIII].  — 

Let  6  =  sin-1  x ;  then  sin  6  =  x,  the  differential  of  which 
by  [XI]  is  cos  0  dd  =  dx; 

dx  dx  dx 


dd  = 


cos  6      Vl  -  sin2  d      Vl  -x< 

dx 


Now  d  cos-1  x  =  d  ( -  —  sin-1  x )  =  — 


78  DIFFERENTIAL  CALCULUS 

57.  Derivation  of  [XIX]  and  [XX].  — 

Let  0  =  tan-1  x;  then  tan  0  =  x,  the  differential  of  which 
by  [XIII]  is  sec2 Odd  =  dx; 

fiR  -     dx  dx  dx 

"  ~  sec2!?  ~~  1  +  tan20  ~  1  +  x2' 

Now  d  cot-1  x  =  d  ( -  —  tan-1  x)  =  —  ..    .     2- 

58.  Derivation  of  [XXI]  and  [XXII].  — 

Let  0  =  sec-1  x;  then  sec  6  =  x,  the  differential  of  which 

by  [XV]  is  sec0tan0d0  =dx; 

...      dd  =        dx        =  dx 

sec  0  tan  0      x  Vsec2  0—1 

dx 


d  cosec-1  x  =  d  f  s  —  sec-1  x)  = 


x  Vx2  -  1 


Now 

\2  I  x  Vx2  -  1 

59.  Derivation  of  [XXIII]  and  [XXIV].  — 

Let  0  =  vers-1  x;  then  vers  d  =  x,  the  differential  of  which 

by  Cor.,  Art.  45,  is  sin0d0  =  dx; 

in  —   dx    _  dx  _ dx 

"  sin0  "  Vl  -  cos2 0  ~  Vl  -  (1  -  vers 0)2 

dx  dx 


Vl-(l-x)2      V2x-x2 

Now  d  covers-1  x  =  d  [  =  —  vers-1  x  )  = . 

\2  /  V2x-x2 


EXERCISE  VI. 


-      ,   •  _,  x  d  (x/a) 

1.   d  sin  *-  —  ■  ,  =  = 


dx 


«      Vl  -  (x/a)2  Va2  -  x2 

,         ,x  — dx  ,,     _.x        adx    . 

2.  d  cos-1  -  =     ,  ;  a  tan  J  -  =    ,,  ,     ., 

a      Va2  —  x2  a      a2  -f  x- 

,  ,x       —adx  ,       _,  x  adx 

d  cot-1  -  =  -=—, — s ;  d  sec 


a      a2  +  x2 '  «      x  Vx2  -  a2 ' 

.         ,  x           —adx  ,         _,  x  dx 

d  esc-1  -  = ,  ;  a  vers  x  -  = 


a      ^  Vx2  -  a2 '  a       V2  ax  -  x2 

Note.  —  These  may  be  considered  standard  formulas. 


EXERCISE  VI 


3.  y  =  tan  x  tan-1  x. 

*     -i     2x 

4.  y=tan»r^ ;• 

5.  ?/  =  sin  l 


15.   ?/  =  tan  1 


16.   ?/  =  arc  sin 


Vl- 
2 


dx 

'  sec2  x  tan-1  x  -\- 

tanx 
1+x2 

dy  _ 
dx 

2  (1  -  x2) 
1  +  6  x2  +  x4 

dy  _ 

1 

V2  dx      Vl-2x-x 


■    ,/-■ —                        dV      Vl  + 
6.   y  =  arc  sin  V  sin  x.  ^-  = ^ 

1.   y  =  xsin_1  x. 


.  cscx. 
dx 


dy 

dx 


=  3*11-1  x  /sin  tg       logx  V 
V    x         Vl-x2/ 


di/  n 

8.  y  =  tan"*  (n  tan  x).  _  -  ____. 

e*  _  e-*  dy  _      -2 

9.  y  =  arc  cos ^j—rx-  ^  -  ^+^' 

2x2  dy  2 

10.  y  =  arc  vers  j-—  ^  =  r+— 2- 

3a; -x3  dy  _      3 

11.  y  =  arc  tan  ^  _  »^» 

12.  j/  ^arcsinj-q^- 

x  +  a 

13.  w  =  arc  tan  ■= 

9  1  —  ax 


dy 
14.    0  =  arc  tan  -j-  • 


dx 

1+x2 

dy  _ 
dx 

-2 

1  +  x2 

dy  _ 
dx 

d<f> 

1 
1+x2 
d*y 

dx2 

dx 
dy 

"Off 

1 

dx 

Vl  -x2 

dy  _ 

-2 

ez  +  e~x  dx.     ex  +  e" 


17.  y  =  arc  tan  (sec  x  +  tan  x).  ~d    =  2 


=  1. 


.    /sinx  — cosx\  dj/ 

18.   y -arc sin ^ ^= J-  ^ 

,3*-2  ,      .  ,3x-12  dy      rt 

A  e^  4-  e_ax  ^  _         2a 

20.   y  =  arc  cot  ^x  _  e-a«'  ^x  "  e20X  +  e^2"*" 


80 


DIFFERENTIAL  CALCULUS 


21.  What  is  the  slope  of  the  curve  y  =  sin  x  ?  Its  inclination  lies 
between  what  values?  What  is  its  inclination  at  x  =0?  What  at 
x=v/2? 

The  slope  =  cos  x;  hence,  at  any  point,  it  must  have  a  value  between 
—  1  and  +1,  inclusive.  Hence,  the  inclination  of  the  curve  at  any  point 
is  between  0  and  7r/4  or  between  3  7r/4  and  r,  inclusive.  (See  figure, 
,Art.  52.) 

60.  Hyperbolic  Functions.  —  These  are  certain  functions, 
recognized  as  far  back  as  1757,  that  have  been  introduced  in 
recent  years,  and  that  are  coming  more  and  more  into  use. 

As  the  trigonometric  functions  are  called  circular  because 
of  their  relation  to  the  circle,  the  hyperbolic  are  so  called 
because  of  their  relation  to  the  rectangular  hyperbola,  the 
relations  being  in  some  respects  the  same.  The  functions 
are  analogous  to  the  trigonometric  functions  and  their  names 
are  the  same.  They  are  the  hyperbolic  sine,  cosine,  tangent, 
etc.,  and  they  are  defined  as  follows: 


sinh  x  =  =  (e2 


e~x),     cschz  = 


1 


coshz 


tanh  x  = 


2  (e*  +  e-»), 

ex  _  e-x 


sechz 


coth  x  = 


sinhz 
1 

cosh  x ' 
ex  +  e~x 


ex  _|_  e-x  ex  _  e- 

61.  General  Relations.  —  Besides  the  reciprocal  rela- 
tions given  above,  the  same  as  those  between  the  circular 
functions,  there  are  analogous  relations : 

cosh2x  —  sinh2  a:  =1;     1  —  tanh2x  =  sech2z; 

coth2  x  —  1  =  csch2  x;    sinh  2  x  =  2  sinh  x  cosh  x; 

cosh  2  x  =  cosh2  x  +  sinh2  x  =  2  cosh2  -1  =  1+2  sinh2  x. 

62.  Numerical  Values.     Graphs.  —  The  sine  may  have 


any  value  from  —  oo  to  oo 


the  cosine  any  value  from  1  to 

/ 


j/=  tanh  j? 


INVERSE  FUNCTIONS  81 

oo  ;  the  tangent  any  value  between  —1  and  1,  and  the  lines 
whose  equations  are  y  =  ±  1  are  asymptotes  to  the  graph  of 
tanh  x.  The  graphs  of  the  sine  and  cosine  are  both  asymp- 
totic to  the  graph  of  y  =  J  ex. 

63.  Derivatives.  —  Since  -j-  e*  =  e*  and  -5-  e-1  =  —  e-1, 

ax  ax 

by  differentiating  the  hyperbolic  functions  as  functions  of  x, 
the  several  derivatives  are  readily  found  to  be  as  follows: 

j-  sinh  x  =  cosh  x;      -7-  cosh  x  =  sinh  x; 

-j- tanh  x  =  sech2x;    -7-cothx  =  —  cosech2x; 

-r-cosechx  =  —  cosechxcothx: 
ax 

-j-  sech  x  =  —  sech  x  tanh  x. 
ax 

The  differentials  are  given  at  once  by  the  derivatives,  or 

vice  versa;  thus,  d  sinh  x  =  cosh  x  dx,  and  so  for  the  others. 

64.  The  Catenary.  —  The  curve  y  =  cosh  x  =  \  (ex+e~x) 
is  called  the  catenary  and  is  important  because  it  is  the 
curve  of  a  perfectly  flexible  and  inextensible  cord  between 
two  points,  and  is  the  curve  that  a  material  cable  when  hung 
between  two  supports  is  assumed  to  take. 

Since     -j-  cosh  x  =  sinh  x,     -p  =  75  (ex  —  e~x)  =  sinh  x 

is  the  slope  of  the  catenary.  The  general  equation  of  the 
catenary  is 

?/  =  a  cosh  -  =  5  \ea  +  e   a/, 

where  a  is  the  distance  from  the  origin  to  the  lowest  point 
of  the  curve.     (See  Art.  146.) 

65.  Inverse  Functions.  —  The  inverse  functions  are 
useful  when  expressed  as  logarithms. 

If  y  =  sinh-1  x,  the  logarithmic  form  of  y  is  found  from 

x  =  sinh?/  =  h(ey  -  e-*), 


82  DIFFERENTIAL  CALCULUS 

which  is  reduced  to 

e2V  -  2xe»  -  1  =  0; 

solving  as  a  quadratic  gives 

ev  =  x  ±  vV  +  1; 

but  as  ey  is  always  positive, 

ev  =  x  +  Vx2  +  1 ;     .*.     sinh-1  x  =  y  =  log  (x  +  Vz2+  1 ) . 
In  the  same  way  is  found,  cosh-1  x  =  log  (x  =b  Vx2  —  1 ) . 

Since  (a  -  V^=l)  -   (aj  +  v?_I)  i 

/.     log  (a  -  Vs2  -  1 )  =  -log  (x  +  Vx2  -  l). 

For  each  value  of  z  greater  than  1  there  are  two  values  of 
cosh-1  x,  equal  numerically  but  of  opposite  sign. 
In  the  same  way  again  is  found, 

1  1  +  x 

tanh"1  x  =  ~  log ,  if  x2  <  1 ; 

COth-1£  =  jrlOff 7,    if    X2  >   1. 

2  x  —  1 

66.   Derivatives  of  Inverse  Functions.  —  The  derivatives 
of  the  inverse  functions  are  found  by  differentiating  their 

logarithmic  forms,  using  formula  -r-  log  x  —  — 

The  derivatives,  taking  -  instead  of  z,  are: 

d    .  .    ,  x  1 

t-  sinh-1 


dx  a      Vx2  +  a2 ' 

d        u  i  x  ! 

-s-  cosh-1  -  =  ±     ,  ; 

dx  a  V^2  -  a2 

3-  tanh-1  -  =  -= r,  (x2  <  a2) ; 

dx  a      a2  —  x1 

-7-  coth-1  -  =  —„ 5,  (x2  >  a2). 

dx  ax2  —  a2 


DERIVATIVES  OF  INVERSE  FUNCTIONS  83 


Since  the  inverse  cosine  is  not  single-valued,  for  the  positive 

ordinate  of  cosh-1  '- 
c 

is  used  instead  of  x, 


X  X 

ordinate  of  cosh-1  L,  the  +  sign  must  be  taken.     When  - 
a  a 


.  ,    ,  x      ,      x  +  Vx-  +  a2 

sinn-1  -  =  log 

a         &  a 

=  log  (x  +  Vx2  +  a2)  —  log  a, 

x 
so  the  derivative  of  sinh-1  -  is  the  same  as  that  of  log 

Qj 

(x  +  Vx2  +  a2),  since  d  (log a)  =  0.     The  divisor  a  occurs  in 

the  logarithmic  form  of  cosh-1  -  also,  so  its  presence  should 

be  borne  in  mind  when  comparing  the  same  result  expressed 
in  logarithms  and  in  inverse  hyperbolic  sines  or  cosines. 

The  relation  of  the  inverse  hyperbolic  sine  to  the  equi- 
lateral hyperbola  is  shown  in  Art.  137,  and  the  inverse  func- 
tions are  again  considered  in  Art.  120. 


CHAPTER  III. 

SUCCESSIVE  DIFFERENTIATION.     ACCELERATION. 
CURVILINEAR   MOTION. 

67.  Successive  Differentials.  —  It  is  often  desired  to 
differentiate  the  differential  of  a  variable  or  to  get  the  deriv- 
ative of  a  derivative.  For,  while  the  differential  of  the  in- 
dependent variable,  being  arbitrary,  is  usually  supposed  to 
have  the  same  value  at  all  values  of  the  variable  and  hence  to 
be  a  constant,  the  differential  of  the  dependent  variable, 
except  when  the  function  is  linear,  is  a  variable,  subject  to 
differentiation. 

The  differential  of  dy  is  called  the  second  differential  of  y; 
the  differential  of  the  second  differential  of  y  is  called  the 
third  differential  of  y;  and  so  on.  d(dy)  is  written  d2y; 
d(d2y)  or  dd  dy,  is  written  d3y;  and  so  on.  The  figure  written 
like  an  exponent  to  d  denotes  how  many  times  in  succession 
the  operation  of  differentiation  has  been  performed,  dy, 
d2y,  d3y,  .  .  .  dny  are  called  the  successive  differentials  of  y. 
Example.  —  The  successive  differentials  of  y  when  y  =  ax3: 
dy  =  3  ax2  dx ; 

d2y  =  Sadx  •  d  (x2)  =  6  ax  dx2; 
d3y  =  Qadx2  •  dx  =  6  a  dx3; 
dAy  =  d  (6  a  dx3)  =  0. 
The  independent  variable  being  x,  dx  is  treated  as  a  constant. 
Note  that  according  to  the  notation  adopted  d2y  =  ddy; 
dy2=(dy)2;    d(y2)  =  2ydy. 

68.  Successive  Derivatives.  —  The  derivative  of  the 
first  derivative  of  a  function  is  called  the  second  derivative  of 
the  function;  the  derivative  of  the  second  derivative  is 
called  the  third  derivative;  and  so  on. 

84 


SUCCESSIVE   DERIVATIVES  85 

When  x  is  independent, 
d  dy  =  d2y      d_  d2y  __  dfy  d   dn~ly  _  dny 

dx  dx  ~  dx2'    dx  dx2  ~  dx3'  '    dx  dxn~l  ~  dxn 

The  successive  derivatives  of  /  (x)  are  denoted  by 

fix),  /"«,  r(x),  r(*)> ■ . .  ,fn(x)> 

Thus  if        /  (x)  =  x\    f  (x)  =  4  x3,    f"  (x)  =  12  x2, 
}"'{x)  =  24  a;,    /IV  (x)  =  24,    /*  (x)  =  0. 

Hence,  if  y  —  J  (x)  and  x  is  independent; 

dx     J  {X)'    dx2     J    Wi  '  '  '  '  dxn     J   {X)' 

The  nth  derivative  of  some  functions  can  be  easily  found  by 
inspection  of  a  few  of  the  derivatives. 

Example  1.—  f(x)  =  ex,f'{x)  =  e*,f"(x)  =  e*,/'"(x)  =  e* 
.  .  .  ,     /.    /» (x)  =  e*. 
This  function  ex  is  remarkable  in  that  its  rate  of  change,  or 
derivative,  is  equal  to  the  function  itself. 

Example  2.-/(0)  =  sin  0,  /' (0)  =  cos  0,  f"  (0)  =  -sin0, 

/'"  (0)  =  -  cos  0,  /IV  (0)  =  sin  0.    /'  (0)  =  cos  0  =  sin  (#  + 1) , 

r  (0)  =  cos  (*  +  !)  =  sin  (*  +  2  •  |) ,    /'"  (0)  =  cos  (0  +  *) 

=  sin(0  +  3.^)  .  .  .  ;     /./"(0)  =  sin(0  +  n.g- 

Each  of  the  successive  derivatives  of  f  (x)  equals  the  x-rate  of 
the  preceding  derivative,  for  fn  (x)  =  -p  fn~l  (x)  =  the  x-rate 

off~*(x). 

Corollary.  —  /n_1  (x)  is  an  increasing  or  a  decreasing 
function  of  x  according  as  fn  (x)  is  positive  or  negative,  and 
conversely. 

Note.  —  The  tangential  acceleration  is, 
_  dv  _  d  ds  _  d2s 
at~~dt~dtdt~M> 


86  DIFFERENTIAL  CALCULUS 

and  the  flexion  is 

,  _  dm  _  d  dy  _  dfy 

dx       dx  dx      dx2 

ds 
(See  Art.  12  and  Art.  13.)     Hence,  the  speed  -r-  is  increasing 

d2s 
or  decreasing  according  as  the  acceleration  -p  is  positive  or 

negative,  and  the  slope  is  increasing  or  decreasing  according 

as  the  flexion  -=2  is  plus  or  minus. 

When  the  second  derivatives  are  equal  to  zero,  the  first 
derivatives  are  constant,  or  conversely.     (See  Art.  13.) 

69.  Resolution  of  Acceleration.  —  An  acceleration,  like 
a  velocity,  being  a  quantity  which  has  magnitude  and 
direction,  may  be  represented  by  a  straight  line,  that  is,  by 
a  vector. 

In  general  the  acceleration  a  at  any  point  (x,  y)  of  a  curvi- 
linear path  may  be  resolved  into  two 
components  in  given  directions.  The 
directions  usually  taken  are  along  the 
tangent  and  normal  at  the  point,  and 
in  directions  parallel  to  rectangular 
axes  OX,  OY.  With  the  notation  of 
the  figure  for  (d),  Art.  10,  the  com- 
ponents parallel  to  the  axes  being  the 
rates  of  change  of  dx/dt  and  dy/dt  will  be  denoted  by.  d2x/dt2 
and  d2y/dt2,  respectively.  The  rate  of  change  of  the  velocity 
is  the  resultant  acceleration  

To  find  the  component  acceleration  at  along  the  tangent 
at  P;  resolve  the  axial  accelerations  along  the  tangent,  giving 
for  the  sum  of  tangential  components, 

d2x  d2y  .  d2x   dx      d-y   dy      d2s 


aI  =  ^cos4>  +  ^sin0=^.^  +  (W2    dg      ^ 


EXERCISE  VII  87 

by  differentiating, 


(dA^/dxY./dyV 
\dt)      \dtj  ^{dtj 


at  =  di?  =  Tt  =  rate  °^  cnan&e  °^  tne  sPeed. 

Hence,  the  tangential  component  is  the  same  as  for  recti- 
linear motion. 

EXERCISE  VII. 

Find  dy,  d2y,  dhj,  when: 

1.  y  =  2  x5  -  5  x4  +  20  x3  -  5  x2  +  2  x. 
dhj  =  120  (x2  -  x  +  1)  dx3. 

2.  y  =  a? log  (x  -  1).  d*y  =  2  ^  ~  3  *  + 3)  dx\ 

(x  -  l)3 

3.  y  =  (x2  -  6  x  +  12)  ex.  d3*/  =  x2e*  dx3. 

4.  i/  =  log  sin  x.  d3?/  =  2  cos  x  sin-3  x  dx3. 

6.   ?/  =  tan  x.  d3?/  =  (6  sec4  x  —  4  sec2  x)  dx3. 

Find  the  successive  derivatives: 

6.  /(x)  =x6  +  4x4  +  3x  +  2.       /VI(s)=[6,    /▼* fcc)  =  0. 
7./  (x)  =  log  (1  +  x);  find  nth  derivative. 
f  (x)  =  (1  +  x)~\    f"  (x)  =  (-1)  (1  +  x)"2, 

f'"(x)    =   (-1)2[2_(1   +X)"3,      /IV(X)    =   (-1)3[3(1  +*)"«,    .    .    . 

.'.      r  (x)  =  (-l)""1  ln-1  (1  +  x)-\ 

8.  /  (x)  =  x3  log  x.  /IV  (x)  =  6  x"1. 

9.  y  =  log  (e*  +  e~x).  p{  =  -8  ,C*  ~  e~*. 
y          &  v               '                         dx6  (ex  +  e-*)3 

10.   Find  formula,  known  as  Leibnitz's  theorem,  for  dn  (uv). 

Let  u  and  v  be  functions  of  x;  then 

d  (uv)  =  dwv  +  u  dv,  (1) 

d2  (uv)  =  d2u  •  v  +  dudv  -\-dudv  -\-u  d2v 

=  d2wv-{-2dudv  -\-  ud2v;  (2) 

.'.     d3  (uv)  =  d*wv  +  3d2udv  +  3 du d2v  +  u d3v.  (3) 

The  coefficients  and  exponents  of  differentiation  are  according  to  the 
Binomial  theorem,  however  far  the  differentiation  is  continued; 

/.     dn  (uv)  =dnu-v  +  ndn~l  udv+  n(wj~  1}  d'^u d2v  +  •  •  . 

+  udu  dn~l  v  +  udnv. 


88 


DIFFERENTIAL  CALCULUS 


11.   Ifz°  +  2,*  =  a°,     g- 


13-  *h+& 


dx2 

1       **- 

'      dx2 


a* 

Z/3 

a2?/3 


12.    If  2/2  =  2px, 


14.    If-„-^  =  l 


6- 


dx2 
d2^ 
dx2' 


y3 

¥ 

a2y3' 


70.  Circular  Motion.  —  When  a  point  describes  a  circle 
of  radius  r  with  constant  speed  v,  it  has  a  constant  accelera- 
tion v2/r  directed  towards  the  center  of  the  circle. 

Let  PT  be  the  velocity  at  P, 
and  P\T\  that  at  P±.  A  velocity 
being  a  directed  quantity  may  be 
represented  by  a  vector;  that  is, 
by  a  straight  line  whose  length 
denotes  magnitude  and  whose 
direction  is  the  given  direction. 
Hence  from  a  common  origin  o, 
the  vectors  op  and  opi  are  drawn 
equal  to  the  vectors  PT  and 
PiTi,  respectively.  Since  the 
speed  is  constant  each  vector  is 
increment,  denoted  by  At;.     The 

Ay 
A* 
directed  along  pph  and  is  laid  off  as  pm. 

As  A£  approaches  zero,  Pi  approaches  P,  and  pi  approaches 
p  along  the  circular  arc  indicated  by  the  dotted  line;  pm 
approaches  a  vector  pt  directed  along  the  tangent  to  the  arc 


v,  and  ppi  is  the  vector 
average  acceleration    for 


the    interval   of   time  A£  is 


the  lim 

A<=0 


m 


represents  the  accelera- 


ppi  at  p.     This  vector, 

dv 
tion  37  of  the  point  P  moving  in  the  circle  of  radius  r;  and 

since  the  direction  is  at  right  angles  with  the  tangent  at  P, 
the  acceleration  is  directed  towards  the  center  0,  is  normal 
acceleration,  therefore,  denoted  by  an.  To  find  the  magni- 
tude of  the  normal  acceleration  an\  since  the  sectors  popi 
and  POPi  are  similar,  the  angles  at  o  and  0  being  equal, 


CIRCULAR  MOTION  89 


arc  ppi      arc  PP\  arc  ppi      As . 

op  OP  v  r 


arc  ppi 


=  ?£,     and    UmP^Ulimrffl 
r  At'  A<-oL     At    ]     r  M^lAtj 

re  ppi  by  its  chord, 

,.      Tchorclppil  _  dv  _  v 

a<=o  L       <^t  dt      r 


At 
replacing  the  arc  ppi  by  its  chord,  (Art.  22.) 

ds 
dt 
dv      v2 

Otherwise,  it  may  be  seen  that  while  the  point  P  describes 
the  circle  of  radius  r,  the  point  p  describes  the  circle  of  radius 
v,  the  velocity  of  p  in  its  path  being  the  acceleration  of  P  in 
its  path.  Since  the  circles  are  described  in  the  same  time, 
the  velocities  are  to  each  other  as  the  paths  of  the  two  points, 
or  as  the  radii  of  the  circles. 

velocity  of  p      velocity  of  P . 
v  r 

velocity  of  p  is  -j ,  rate  of  change  of  velocity  v  of  P, 

an      v  v2 

—  =  -  i     an  =  —  • 
v       r  r 

Since  the  speed  is  constant,  the  rate  of  change  at  is  zero, 

v" 

cit2  +  an2  =  an  =  - ;  that  is,  the  total  acceleration 
r 

is  the  normal  acceleration,  the  change  of  velocity  being  change 

of  direction  only.     Since 

4?rV 
s  =  2irr  =  vt;     a  =  -^-,  (2) 

where  T  is  time  of  a  revolution. 

Note.  —  By  Newton's  Second  Law  the  measure  of  the 

W 

force  on  a  moving  body  is  —  a  (Art.  71);  hence,  the  force 

acting  on  a  body  weighing  W  lbs.  revolving  in  a  circle  of 


90  DIFFERENTIAL  CALCULUS 

Wv2 

radius  r  is    pounds  of  force,  is  directed  towards  the 

gr 

center  of  the  circle,   and  is  called  centripetal  force.     The 

reaction  of  the  body  to  this  force  is  by  the  Third  Law  equal 

in  magnitude  and  opposite  in  direction.     It  acts  upon  the 

axis  or  upon  whatever  deflects  the  body  from  its  otherwise 

rectilinear  path,  and  has  been  called  the  centrifugal  force, 

although  a  misnomer.     The  centripetal  force  is  the  active 

force,  the  other  is  the  equal  and  opposite  reaction  and  should 

be  called  the  centrifugal  reaction,  since  it  is  the  resistance 

which  the  inertia  of  the  body  opposes  to  the  force  acting 

upon  it. 

71.  The  Second  Law  of  Motion.  —  According  to  New- 
ton's Second  Law  of  Motion  the  rate  of  change  of  momentum 
of  a  body  is  proportional  to  the  resultant  of  the  impressed 
forces  acting  on  the  body. 

Let  a  body  of  standard  weight  W  be  moving  with  velocity 
v,  then  Wv  =  momentum  of  the  body; 

•      <L  (Wv)  =  W—  =  W—  =  Wa, 
"     dtKVVV)      W  dt      W  dt2      wat' 

Hence,  if  F  be  the  resultant  force  and  a  the  acceleration,  by 
the  Law, 

Wa<xF    or    Wa  =  kF;  (1) 

that  is,  the  product  of  the  numbers  representing  the  weight 
and  the  acceleration  is  proportional  to  the  number  represent- 
ing the  force.  The  value  of  the  factor  k  depends  upon  the 
units  used  for  the  other  factors.  When  these  are  the  usual 
units,  foot,  pound  (weight),  second,  pound  (force),  it  is 
found  by  experiment  that  k  has  the  value  32,  approximately. 
Experiment  shows  that,  while  the  value  varies  slightly  for 
different  localities,  it  is  the  same  for  all  bodies  in  any  one 
locality.  This  value  is  denoted  by  g  and  is  called  the  accel- 
eration of  gravity;  for  when  a  body  falls  freely,  gravity  being 
the  only  force  acting,  the  acceleration  is  found  to  be  about 


THE   SECOND  LAW  OF  MOTION  91 

32  feet  per  sec.  per  sec.  The  locality  in  which  g  =  g0  = 
32.1740  ft. /sec.2  has  been  adopted  as  the  "  standard  locality  " 
and  the  weight  of  the  body  in  that  locality  is  called  the 
standard  weight  of  the  body.  Putting  g  for  k  in  equation  (1) 
it  becomes 

W 
Wa  =  gF    or    F  =  —a.  (2) 

If  the  force  F  is  the  force  of  gravity  acting  on  W,  then 
dv 

di  =  a  =  g- 

the  acceleration  of  gravity;  for  tip  weight  W  is  the  force  of 
gravity  acting  on  the  body  denoted  by  the  letter  W. 

Since  for  any  given  body  the  ratio  of  the  force  to  the 
acceleration  produced  is  constant,  the  value  of  this  ratio, 
F/a  or  W/g,  is  a  characteristic  of  the  body,  called  its  inertia 
and  the  ratio  may  be  denoted  by  the  letter  m;  then  the 
equation  (2)  may  be  written 

F  =  ma.  (3) 

In  using  equation  (3)  for  the  solution  of  problems,  with  the 
usual  units,  m  must  be  replaced  by  W/g. 

Some  writers  use  the  word  "mass"  to  denote  the  inertia, 
while  others  use  it  for  standard  weight;  consequently,  there 
are  some  who  avoid  the  use  of  the  word  on  account  of  the 
resulting  confusion. 

In  Physics  the  equation  (3)  is  used,  the  unit  of  force,  called 
the  absolute  unit,  being  that  unit  which  in  equation  (1) 
makes  k  =  1,  the  other  units  remaining  the  same,  and, 
therefore,  m  measured  in  pounds  the  same  numerically  as 
the  standard  weight  W. 

Accordingly;  if  the  number  g  be  the  number  of  absolute 
units  of  force  with  which  gravity  attracts  the  unit  mass  (or 
weight),  the  Law  becomes 

mrit=  m^J     nence     ~Ji  =  9>  tne  acceleration  of  gravity. 


92  DIFFERENTIAL  CALCULUS 

The  absolute  unit  of  force  is  thus,  that  force,  which  acting  on 
the  unit  of  mass  (or  weight)  for  the  unit  of  time,  generates  the 
unit  of  velocity.  The  absolute  unit  of  force  is  thus  1/g  of  a 
pound  avoirdupois,  about  \  of  an  ounce,  and  F  is  given  in 
this  unit  when  in 

F  =  ma, 

m  is  expressed  in  pounds,  the  unit  being  a  pound.  The 
ordinary  unit  of  force,  sometimes  called  the  Engineer's  unit, 
is  one  pound  and  is  g  times  the  absolute  unit  used  in  Physics. 
Newton's  Second  Law  of  Motion  gives  as  a  definition  of 
force:  force  is  the  time-rate  of  change  of  momentum.  Using 
the  much  abused  term  "mass,"  the  definition  is:  the  force 
is  the  product  of  the  mass  times  the  acceleration.     From 

W 

F  =  ^a;  (2) 


9 

Wd?s  Wv* 

6   dt*>  tn       gr 


w       WdH  W<Py 

tx~  g    dP'  tv~  g    dP' 

the  tangential,  normal,  and  axial  components  of  a  force  F, 
corresponding  to  the  accelerations,  at,  a„,  ax,  ay.  Since 
kinetic  energy  of  a  moving  body  is  E  =  \  mv2, 

dE 

-j-  =  mv, 
dv 

that  is,  the  i>-rate  of  E  is  momentum; 
dE  dv       „ 

that  is,  the  time-rate  of  E  is  product  of  force  and  velocity. 

72.  Angular  Velocity  and  Acceleration.  —  When  a  body 
is  rotating  about  an  axis  the  amount  of  rotation  depends 
upon  the  time;  so  if  6  is  the  angle  through  which  any  line  in 
the  body,  intersecting  the  axis  at  right  angles,  turns,  then  0 


ANGULAR  VELOCITY  AND  ACCELERATION  93 

gives  the  amount  of  rotation  and  is  a  function  of  the  time  t. 
Thus  in  the  case  of  a  wheel  the  rotation  is  measured  by  the 
angle  0  through  which  a  spoke  turns  in  a  time  t.  The  rota- 
tion is  uniform  if  the  bod}-  rotates  through  equal  angles  in 
equal  intervals  of  time.  The  rate  of  rotation  or  the  rate  of 
change  of  the  angle  is  the  angular  velocity  or  speed  and  is 
denoted  by  co. 

If  the  rotation  is  uniform,  the  angular  velocity  is  constant 

a 

and  co  =  - ,  9  being  in  radians ;  hence,  if  the  uniform  rate  of 

rotation  is  co  radians  per  second,  the  body  rotates  through  cot 
radians  in  t  seconds  of  time. 

If  the  rotation  is  not  uniform  the  rate  at  which  the  body 
is  rotating  at  any  instant  is  the  angular  velocity  at  that  time, 

.  ..     A0      dd 

and  co  =  lrm  —  =  -=r  • 

a<=o  &      dt 

This  expression  for  angular  velocity  is  general  and  is  applic- 
able when  the  rotation  is  uniform  also;  for  then, 

_  6  _  Id  _  dd 
W  ~  t  ~  \t  ~  dt' 


although,  the  ratios  being  constant,  no  limit  is  involved 

Similarly,  the  a 
acceleration ;    and 


Similarly,  the  angular  acceleration  is  a  =  - ,  for  constant 


=  dco  =  d  (d&\  =  <Pd 
a~  dt  ~  dt[dt)  ~  dt2 


is,  in  general,  the  time-rate  of  change  of  angular  velocity, 
or  the  angular  acceleration. 

If  a  particle  is  at  a  distance  r  from  the  axis  of  rotation,  the 
relation  between  the  angular  velocity  of  the  particle  and  its 
linear  velocity  follows  at  once,  whether  the  rotation  is  uni- 
form or  not. 


94 


DIFFERENTIAL  CALCULUS 


Since  in  circle 

As  =  r  •  AO, 

Ad  _  1     As 
M      r'  At' 

..     A0      1    ..     As 
lim  -t—  =  -  •  lim  77 . 

At=o  At      r     A<=o  A^ 

»*■ 

dd      1     ds      v 
"  ~  dt  ~  r  '  di  ~  r' 

the  relation  sought.  Hence,  since 
the  angular  velocity  of  every  point 
of  a  rotating  body  has  the  same 
value  at  any  instant,  and  the  direc- 
tion of  motion  of  a  particle  at  any 
point  is  along  the  tangent  at  the 
point, 

tangential  velocity  v  =  rco, 
if  co  is  the  angular  velocity  of  the 
"particle  about  axis  at  0. 


Since 


dv 
dt 


d{ru>) 
dt 


or 


d?s=  m 

dt2      rdFi 


tangential  acceleration  at  =  ra, 

the  relation  between  tangential  acceleration  and  angular 
acceleration  when  a  is  the  angular  acceleration  of  a  particle 
at  a  distance  r  from  the  axis  of  rotation. 
dv 


2,.,2 


Since 


=  ro)- 


dt~r 
normal  acceleration  an  =  rar. 


73.  Simple  Harmonic  Motion.  —  If  a  point  move  uni- 
formly on  a  circle  and  the  point  be  projected  on  any  straight 
line  in  the  plane  of  the  circle,  the  back-and-forth  motion  of 
the  projected  point  on  the  given  straight  line  is  called  simple 
harmonic  motion.  It  is  denoted  by  the  letters  S.  H.  M. 
Let  the  point  P  move  upon  the  circumference  of  a  circle  of 
radius  a  with  the  uniform  velocity  of  v  feet  per  second,  so 


SIMPLE   HARMONIC   MOTION 


95 


that  the  radius  OP  rotates  with  uniform  angular  velocity  at 

the  rate  of  -  =  co  radians  per  second.     The  projection,  P', 
a 

of  P  on  the  vertical  diameter,  moves 

up  and  down.     Let  6  be  the  angle  that 

the  radius  makes  with  the  z-axis,  then 

if  the  point  P  was  at  A  when  t  =  0,  the 

displacement  OP'  =  y  is  given  by 

y  =  a  sin  6  =  a  sin  ut. 

If  the  point  P  was  at  P0  when  t  =  0,  and  at  A  when  t  =  t0, 

vt 
then         y  =  a  sin  (ut  —  a),     where     a  =  ut0  =  — •  (1) 


When  the  displacement  at  time  t  is  given  by  (1)  the  motion 
is  S.  H.  M.     Hence,  the  point  P'  describes  S.  H.  M. 

The  velocity  of  a  point  describing  S.  H.  M.  is,  from  (1), 


dy  ,  ±        x 

-j7  =  aw  cos  (cot  —  a) , 


and  the  acceleration  is 


d2y  2   .    ,  ,  •     N 

-~nf2  ~  ~ aoi  sm  (^ —  a) 

=  -uhj,  from  (1), 


or 


d2y 


dP 


l  +  ^y  =  0. 


(2) 

(3) 
(4) 
(5) 


It  should  be  noted  that  equation  (1)  may  be  written  in  the 
form 

y  =  a  sin  (cot  —  a)  =  a  [sin  ut  cos  a  —  cos  wt  sin  a] 
or  y  =  A  sin  kt  +  B  cos  kt, 

where  A  =  a  cos  a  and  B  =  —  a  sin  a  are  constants.  This 
equation  and  y  =  a  sin  (A;£  —  a)  are  the  general  formulas 
for  S.  H.  M. 

The  acceleration  of  a  particle  describing  S.  H.  M.,  as 
shown   by    (4),   is   proportional   to   the   displacement   and 


96 


DIFFERENTIAL  CALCULUS 


oppositely  directed.  It  is  oppositely  directed  since  the 
motion  is  one  of  oscillation  about  a  position  of  equilibrium. 
When  the  body  is  above  this  position  the  force  is  directed 
downward,  and  when  it  is  below,  the  force  is  upward.  In  the 
figure  the  point  P'  has  a  negative  acceleration  when  above  0 
and  a  positive  acceleration  when  below  0.  The  acceleration 
is  zero  at  0,  a  maximum  at  B'  and  a  minimum  at  B;  while 
the  corresponding  velocity,  as  given  by  (2) ,  has  its  maximum 
numerical  value  as  P'  passes  through  0  in  either  direction, 
and  is  zero  at  B  and  B' ',  the  ends  of  the  vertical  diameter. 
The  factor  of  proportionality  or  is  connected  with  the  period 

2  7T 

T  by  the  relation  T  =  — ,  where  the  period  of  the  S.  H.  M. 

CO 

y  =  a  sin  ait  is  the  time  T  required  for  a  complete  revolution 


of  the  point  P;    that  is,  wT  =  2t.     The  time   t0  = 


to 


make  part  of  a  revolution  is  called  the  phase,  a  being  epoch 
angle.        The  number  of  complete  periods  per  unit  of  time 

is  N  =  7„  =  ?r~ >  where  N  is  the  frequency  of  the  S.  H.  M. 

Let  P'  be  a  tracing  point  capable  of  describing  a  curve  on  a 
uniformly  translated  sheet  of  paper,  SS',  then  if  the  sheet  be 


moved  with  the  same  speed  as  the  point  P  moves  on  the 
circumference  of  the  circle  of  radius  a,  P'  describing  S.  H.  M. 
on  the  vertical  diameter  will  trace  the  sinusoid  P'BP'B'  on 
the  moving  paper.     The  sinusoid  will  have  as  its  equation 


v  x 

y  —  a  sin  - 1  =  a  sin  - , 
a  a 


where  x  is  the  abscissa  of  any  point  of  the  sinusoid  referred 


EXERCISE   VIII  97 

to  an  origin  (as  (V)  moving  with  the  paper.  The  circle  is 
shown  in  the  figure  in  several  positions  corresponding  to  the 
different  angles  through  which  the  radius  OP  has  revolved, 
or  the  different  positions  of  the  projected  point  P'  on  the 
vertical  diameter  BOB'.  The  amplitude  of  the  S.  H.  M.  is 
the  same  as  that  of  the  sinusoid;  that  is,  the  radius  a  of  the 
circle.     The  period  of  the  sinusoid  is  2  wa,  corresponding  to 

the  period,  T  =  — ,  of  the  S.  H.  M.  of  the  point  P'  on  the 

CO 

vertical  diameter. 

74.  Self -registering  Tide  Gauge.  —  The  principle  by 
which  the  up-and-down  motion  of  a  point  is  represented  by 
a  curve  is  utilized  in  the  self -registering  tide  gauge  for  record- 
ing the  rise  and  fall  of  the  tide.  Such  a  gauge  consists 
essentially  of  a  float  protected  by  a  surrounding  house  or 
tube,  and  attached  by  suitable  mechanism  to  a  pencil  that 
has  a  motion  proportional  to  the  vertical  rise  and  fall  of  the 
float.  The  pencil  bears  against  a  piece  of  graduated  paper 
fastened  to  a  drum  that  is  revolved  by  clockwork.  There 
will  thus  be  drawn  on  the  paper  a  curve  where  the  horizontal 
units  are  time,  and  the  vertical  units  are  feet  of  rise  and  fall. 
The  stage  of  the  tide  is  given  for  any  time. 

EXERCISE  Vm. 

1.  The  angle  (in  radians)  through  which  a  rotating  body  turns, 
starting  from  rest,  is  given  by  the  equation 

6  =  |  at0-  +  u0t  +  <?o, 

where  a,  w0,  60  are  constants;  find  the  formulas  for  angular  velocity  and 
angular  acceleration  after  any  time  t. 

w  =  -7-  =  at  +  wo,  which  gives  the  angular  velocity; 

tt  =  -jEjr  =  a,  which  gives  the  angular  acceleration. 

2.  A  flywheel  is  brought  from  rest  up  to  a  speed  of  60  revolutions 
per  minute  in  \  minute.     Find  the  average  angular  acceleration  a,  and 


98  DIFFERENTIAL  CALCULUS 

the  number  of  revolutions  required.     Find  the  velocity  at  the  end  of 
15  seconds. 

to  =  60r.p.m.  =  60  X  ~tt  =  2  tt  radians  per  sec. 
/.      at  =  a  •  30  =  2tt    or    a  =  ~  =  0.2094  rad./sec2. 

6  =  \  at2  =  \  M  (30)2  =  15  X  2tt  =  15  revolutions. 
co  =  at  =  —  X15=7r  =  3.14  rad.  per  sec. 

3.  If  the  flywheel  of  Ex.  2  is  12  feet  in  diameter,  find  the  tangential 
velocity  and  acceleration  of  a  point  on  the  rim.  Find  the  normal 
acceleration  at  the  instant  full  speed  is  attained. 

y  =  rco  =  6X27r  =  37.7  ft.  per  sec. 
af=ra  =  6x|^  =  6X  0.2094  =  1.256  ft. /sec2. 


CHAPTER  IV. 


GEOMETRICAL  AND  MECHANICAL  APPLICATIONS. 


dy 
dx 


75.    (a)  Tangents  and  Normals.  —  Since  the  derivative 
=  /'  (x)  represents  the  slope  of  the  curve  y  =  fix)  at  any 


point  0,  y), 


[21 = siope 


of  PiT  =  tan  0  =  mi, 


where  4>  is  the  angle  XTQ,  measured  from  the  positive 
direction  of  the  z-axis  to  the  tangent  TPh  and  mi  is  the 
slope  of  the  curve  at  the  point  Pi  (xh  yi) . 

Hence  from  the  equation  of  a  line  through  a  given  point 
(xh  Vi)i  V  —  2/i  =  wi(£  —  Xi);    the  equation  of  the  tangent 

at  the  point  Pi  (xh  yi)  is  y  —  yx  =  \-r\  (x  —  Xi),  in  which  the 

subscript  denotes  that  the  quan- 
tity is   taken   with   the  value 
which  it  has  at  the  point  Pi. 
Since  the  sign  of  the  derivative 
of  a  function  indicates  whether 
the   function    is    increasing   or 
decreasing,  when  mi  is  positive 
the  curve  is  rising  at  Pi,  and 
when  mi  is  negative  the  curve 
is  falling  there.     If  mi  is  zero 
the  tangent  is  horizontal,  parallel  to,  or  coincident  with  the 
z-axis;   and  if  mi  is  infinite,  the  tangent  is  vertical,  parallel 
to,  or  coincident  with  ?/-axis. 
Points  where  the  slope  has  any  desired  value  can  be  found 

99 


Tii  = =  —  cot  A 

mi 


100  DIFFERENTIAL  CALCULUS 

by  setting  the  derivative  equal  to  the  given  number  and 
solving  the  resulting  equation  for  x. 

The  slope  of  the  normal  NPh  being  the  negative  reciprocal 
of  the  slope  of  the  tangent  TPh  is 

\  _dx\ 

L     dy\; 
Hence,  the  equation  of  the  normal  is 

=[~dil(x~Xi)  or  x-Xi+iy-yi)Mr°- 

(b)  Subtangents  and  Subnormals.  —  The  subtangent 
and  the  subnormal  are  the  projections  on  the  z-axis  of  the 
part  of  the  tangent  and  normal,  respectively,  between  the 
point  of  tangency  and  the  x-axis. 

From  the  figure: 

dx 
Subtangent  TM  =  yi  cot  4>  =  y\ 


y-yi 


Subnormal  MN  =  yi  tan  <j>  =  y± 


dy. 
'dy 
dx 


Tangent       TP,  =  VmP*  +  TM*  =  \/y^  +  yf  [j|  J 


2/i  V  1  +  hr     =2/i cosec  <£• 


Normal        NPX  =  VmP*  +  MN2  =  \J  y?  +  tf  \^-~f 

=  2/iVl+[JJ1  =  2/isec</). 

If  the  subtangent  is  reckoned  from  the  point  T,  and  the 
subnormal  from  the  point  M,  each  will  be  positive  or  nega- 
tive according  as  it  extends  to  the  right  or  to  the  left.  For 
any  given  curve  the  signs  will  depend  upon  the  coordinates 
of  the  point  of  tangency. 

Note.  —  As  mentioned  before,  the  problem  of  tangents 
directly  led  to  the  Differential  Calculus. 


r*i- 


ILLUSTRATIVE  EXAMPLES  101 

76.   Illustrative  Examples.  —  1.  The  circle  x2  +  y2  =  a2. 
Differentiating,         2  x  dx  +  2  y  dy  =  0, 

ty  -  —- 
dx  y 

yi 

Equation  of  tangent, 

y-yi=  -~-(x-+xi) 

or  xxi  +  yyi  =  a2,  after  reducing. 
Equation  of  normal, 

y  -  yi  =  z-  (*  -  a?i) 

or  yxi  —  y\X  =  0,  after  reducing. 

The  final  form  of  the  last  equation  shows  that  the  normal 
at  any  point  on  the  circle  passes  through  the  center. 

The  subtangent 

™-»ia-»(-s--ff— H*- 

The  subnormal    MN  =  jfrD^l  =yJ-^\=  -Xl. 

x                                                                        x 
Since  dy  = dx,  the  ordinate  of  the  circle  changes 

y  y 

x 
times  as  fast  as  the  abscissa ;  and  since  dy  — dx  is  nega- 

y 

tive,  unless  x  and  y  have  different  signs,  y  is  a  decreasing 
function  of  x  in  the  first  and  third  quadrants;  while  dy  being 
positive  when  the  moving  point  is  generating  the  second  and 
fourth  quadrants,  y  is  an  increasing  function  of  x  in  those 
quadrants. 

2.   The  parabola  y2  =  2  px. 

Differentiating,  2ydy  =  2pdx, 

.      dy 
dx 


2,        •     [dy]  =  R, 

y  "     [dxji      ?/i 


102  ^DIFFERENTIAL  CALCULUS 

Equation  of  tangent,  y  —  yi  =  —  (x  —  Xi)  or  2/2/1  =  p(x  +  Xi), 
after  reducing. 

Equation  of  normal,  y  —  yi  =  —  —  (x  —  xi). 
The  subtangent  TM  =  yJ^j  =  Vl  =  ?£*  =2Xl. 
'The  subnormal  MT  =  yj— )  =  p. 

Hence,  for  any  point  on  the  parabola  the  subtangent  is 
bisected  at  the  vertex  and  the  subnormal  is  constantly  equal 
to  p,  the  semi-latus  rectum.  These  two  characteristics  of 
the  parabola  afford  ready  methods  of  accurately  drawing  a 
tangent  at  any  point  on  the  curve. 

Since  dy  =  -  dx,  the  rate  of  y  =  -  times  rate  of  x.    To 

y  y 

find  where  the  rates  are  the  same,  put  ~  =  -  =  1,   /.    y  =  p 

(xx      y 

and  x  =  ^ ;  that  is,  the  extremity  of  the  latus  rectum  is  the 

point  where  the  rates  of  y  and  x  are  equal.  Hence  the  tan- 
gents at  the  extremities  of  the  latus  rectum  make  angles  of 
45°  and  135°  with  z-axis  and  meet  at  right  angles  with  each 
other  at  intersection  of  directrix  and  a>axis.     It  is  evident 

dy 
that  at  the  origin  where  y  =  0,  -~  =  oo  ;  that  is,  the  y-axis 

is  tangent  at  the  vertex.  It  is  seen  also  that,  as  y  increases 
without  limit,  the  tangent  at  its  extremity  becomes  more 
and  more  nearly  parallel  to  the  x-axis. 

3.  On  the  circle  x2  +  y2  =  1,  to  find  the  points  where  the 
slope  is  1,  0,  or  oo . 


r^/i  =  _  *1  =  i    ... 

\_dxji  2/1 


2/i  =  -xi; 


ILLUSTRATIVE  EXMAPLES  103 

substituting  in  x2  +  2/2  =  1,  Xi  =  =b£  V2  and  2/1  =  =F=J  V2. 

B1--2-*  •'•  XI  =  0  and  *=±1' 

by  substituting  in  x2  +  2/2  =  1- 

[21=-|=go'  •■•  »-°  and  *-±i' 

by  substituting  in  x2  +  ?/2  =  1. 

4.  To  find  at  what  angle  the  circle  x2  +  y2  =  8  and  the 
parabola  y2  =  2x  intersect. 

Making  the  two  equations  simultaneous,  the  points  of 
intersection  are  found  to  be  (2,  2)  and  (2,  —2). 

=Fl. 


For  circle,       mi  = 

~d?/~ 

xi  _         2 

_dx_ 

i "      yi  "       ±2 

For  parabola,  mi  = 

~dy~ 
_dx_ 

i      2/i       ±2 

For  angle  of  intersection, 

2          1 

.               mi- 
tan  4>  =  r— — 

-  m 
mirr 

2            ±2       d=2 

2                     1-i 

=  =F3, 

or  <f>  =  tan"1  (=F3),  and  from  table  of  tangents,  0  =  108°  26' 
or  71°  34'. 

5.  The  path  of  a  point  is  the  arc  of  a  parabola  y2  =  2  px, 
and  its  velocity  is  v;  find  its  velocity  parallel  to  each  axis. 

Let  s  denote  the  length  of  the  path  measured  from  any 

point  on  it;  then  -r-  =  v. 

From  y2  =  2  px, 

dy  _  p  dx 

dt      y  dt 
Substituting  these  values  in 


(Shift' HW  <-•»». 

\dt)  +y*Ut)' 


104  DIFFERENTIAL  CALCULUS 

/dx\2  _      v2  y2v2  dx  _         yv 


IT 


eft      i/  oft       Vw2  +  p2 

6.  A  comet's  orbit  is  a  parabola,  and  its  velocity  is  v;  find 
its  rate  of  approach  to  the  sun,  which  is  at  the  focus  of  its 
orbit. 

Let  p  denote  the  distance  from  the  focus  to  any  point  on 
y2  =  2px;  then  p  =  x  +  \  p,  from  point  to  directrix; 

dp  _  dx 
"     di~di' 

p  being  constant.  Hence,  the  comet  approaches  or  recedes 
from  the  sun  just  as  fast  as  it  moves  parallel  to  the  axis  of 
its  orbit; 

dp      dx  y 


dt        dt        \/y1 


v.     (Example  5.) 


P 


At  the  vertex,  y  =  0;  hence,  at  the  vertex  -7-  is  zero.     When 
y  =  V, 


<1. 


and 


dp 

It  ~ 

=  |V2, 

dp 

dx 
~dt 

<  v,     s 

y 

dt 

Vy2  +  p2 

dy  _ 
dt 

pv 
Vy2  +  p2 

>    (Example  5) 

lim  —7 

pv 

.        -0: 

dx  .  ds 

-di  =  dt=V>     SmCe 


(ds\2  =  (dx\\  (dy\* 
\dt)  "  \dt)   r\dt) 

hi 

limit  as  y  increases  without  limit. 


Hence  -r-  =  -=7  is  always  less  than  v  and  approaches  v  as  a 


ILLUSTRATIVE  EXAMPLES 


105 


7.  To  compare  the  velocity  of  a  train  moving  along  a 
horizontal  tangent  with  the  velocity  of  a  point  on  the  flange 
of  one  of  the  wheels,  and  to  compare  also  the  horizontal  and 
vertical  components  of  the  flange  point. 

Let  a  wheel  whose  radius  is  a  roll  along  a  horizontal  line 
with  a  velocity  v\  find  the  velocity  of  any  point  P  on  its  rim, 
also  the  velocity  of  P  horizontally  and  vertically. 


)  =  a  vers  6,    ) 


(i) 


The  path  of  P  is  a  cycloid  whose  equations  are: 
x  =  a  (6  —  sin 
y  =  a  (1  —  cos0) 

where  6  denotes  the  variable  angle  DCP,  and  a  the  radius 
CD. 

Since  the  center  of  the  wheel  is  vertically  over  D, 

v  =  the  time-rate  of  OD 


d  (ad) 


di 


dd 
dt' 


cW 
dt 


(2) 


Differentiating  equations  (1)  gives,  by  (2), 

v  vers  6 
=  the  velocity  horizontally, 


dx         ,  dO  adB 

37  =  a  (1  —  cos  6)  -r±  =  a  vers  0  -y- 
at  at  at 


and 


dy  .      dd 

f-  =  a  sin  0  jt 
dt  dt 


v  sin  6  =  the  velocity  vertically. 


(3) 
(4) 


106  DIFFERENTIAL  CALCULUS 


=  velocity  of  P  along  its  path.      (5) 

The  velocity  of  P  may  be  considered  as  the  resultant  of  two 
velocities  each  =  v,  one  along  PT  tangent  to  the  circle  and 
the  other  along  PH  parallel  to  the  path  of  C.  The  resultant 
PB  must  bisect  the  angle  HPT;  :.  DPB  =  90°  and  PB  is 
tangent   to   the   cycloid,  the   path  of  P,  making  PD  the 


ds 

di  =  V' 


normal. 

At  0, 

e 

=  o: 

MP, 

e 

"  3 

At  Pi, 

,e 

7T 

~  2 

d 

dx      dy      ds  _  ft 
dt       dt      dt 

dx      1         dy      1      /~ 
Tt=2V'     J  =  2^ 

dx      dy             ds           /x 

Tt  =  Tt  =  v>    di  =  vV2' 

tk-^-2v       dy-0 
dt      dt           '      dt 

A 

ds 
di 

:  v  =  V2a-y  :  a. 

At  P2,  9  =  ir, 
From  (5)  is  obtained 


Hence,  the  velocity  of  P  is  to  that  of  C  as  the  chord  DP  is 
to  the  radius  DC;  that  is,  P  and  C  are  momentarily  moving 
about  D  with  equal  angular  velocities.     (See  Art.  72.) 

When  6  =  60°,  their  linear  velocities  also  are  equal,  as 
shown  above. 

8.  Find  the  equation  of  the  tangent  and  the  Values  of  the 
subnormal  and  normal  of  the  cycloid. 

Dividing  (4)  by  (3),  Example  7,  gives 


sin0 

V(2 

a  —  3 

y)y/a> 

vers  6 
PH  _ 

VHB- 

1)11 

and 

dy  =   sinfl  =  V(2a  -  y)y/a  =  . /(2a-  y) 
dx      vers  6  y/a  V         y 


sin  6  =  -777S  = and     vers  6  =  -  from  (1) ; 

CP  a  a 


EXERCISE  IX  107 

is  the  equation  of  the  tangent  at  point  (xh  yi). 

m        i  i        dy         sin  6         sin0  .    _     „„     ,,_ 

The  subnormal  =  y-r  =y z  =  y  — j-  =  a  sin  0  =  PH  =  MD. 

J  dx     y  vers  B     9  y/a 

Thus  the  normal  at  P  passes  through  the  foot  of  the  per- 
pendicular to  OX  from  C.  Hence,  to  draw  a  tangent  and 
normal  at  P,  locate  C,  draw  the  perpendicular  DCB  equal 
to  2  a,  and  join  P  with  B  and  D;  then  PB  and  PD  will  be 
respectively  the  tangent  and  normal  at  P. 

Normal  =  DP  =  VDB  •  DH  =  V2a-y. 

9.   Eliminating  0  in  equations  (1)  of  Example  7 ,  equation 
of  cycloid  is 

x  =  a  •  arc  vers  y/a  =F  V2  ay  —  y~, 
since     0  =  arc  vers  y/a  and  a  •  sin  0  =  d=  V(2  a  —  ?/)  a. 


EXERCISE   IX. 

Deduce  the  following  equations  of  the  tangent  and  the  normal: 

1.  The  ellipse,  x2/a2  +  y2  IIP-  =  1,     xix/a2  +  yiy/b2  =  1, 

2.  The  hyperbola,  x2/a2  -  y2/b2  =  1,     Xix/a2  -  yiy/b2  =  1, 

y-y,  =  -^{x-  x:). 

3.  The  hyperbola,  2xy  =  a2,     Xi?/  +  y\X  =  a2,     yxy  —  X\X  =  yi2  —  Xi2. 

4.  The  circle,  x2  +  y2  =  2  ax,     y  —  y%  =  {x  —  Xi)  (a  —  Xi)/yh 

y  —  yi  =  (x-  xi)  yi/(xi  -  r). 

5.  Find  the  equations  of  the  tangent  and  normal  at  (3/2 a,  3/2 a): 
x3  -f  y3  =  3  ax?/.  A?is.  x  +  ?/  =  3a,     x  =  ?/. 

6.  x  +  ?/  =  2c*-*  at  (1,  1). 

Ans.  3y  =  x  +  2,  3  x  +  y  =  4. 

7.  (x/a)n  +  (y/b)n  =  2,  at  (a,  6). 

Ans.  x/a  +  ?//&  =  2,  ax  —  by  =  a2  —  b2. 


108  DIFFERENTIAL  CALCULUS 

8.  Show  that  the  sum  of  the  intercepts  of  the  tangent  to  the  para- 
bola x*  +  V*  =  a*,  is  equal  to  a. 

9.  Show  that  the  area  of  the  triangle  intercepted  from  the  co- 
ordinate axes  by  the  tangent  to  the  hyperbola,  2  xy  =  a2,  is  equal  to  a2. 

10.  Show  that  the  part  of  the  tangent  to  the  hypocycloid  x3  +y *  =  a *, 
intercepted  between  the  axes,  is  equal  to  a. 

11.  Find  the  slope  of  the  logarithmic  curve  x  =  log6  y.     The  slope 
varies  as  what  ?     What  is  the  slope  of  the  curve  x  =  log  y  ? 

12.  Find  the  normal,  subnormal,  tangent,  and  subtangent  of  the 
catenary  y  =  a/2  (e*/a  +  e~x'a). 

Ans.  y2/a;   a/4  (e2X/a  -  e-2X'a):    ■      ^        \       ,  ay 

Vy*  -  a2      V2/2  -  a2 

13.  At  what  angles  does  the  line  Sy  —  2x  —  8  =  0  cut  the  parabola 
=  8  x?  Ans.  arc  tan  0.2;     arc  tan  0.125. 

77.   Polar  Subtangent,  Subnormal,  Tangent,  Normal.  — 
Let  arc  mP  =  s,  and  arc  PQ  =  As;    then    £  POQ  =  Ad, 


T 
circular  arc  PM  =  p&6,  and  MQ  =  A  p.     The  chords  PM 
and  PQ,  the  tangents  RPH  and  TPZ,  are  drawn;   and  ZH 
is  drawn  perpendicular  to  PH,  Z  being  any  point  on  the 
tangent  PZ. 

When  As  =  0,  the  limiting  positions  of  the  secants  PM 
and  PQ  are  the  tangents  RPH  and  TPZ,  respectively;  hence, 
It  ( Z  PMQ)  =  /  RPK  =  tt/2  =  z  PHZ, 
U(Z  OQP)  =  z  OPT  =  +  =  z  #ZP, 
and  It  ZMPQ=  zHPZ. 


POLAR  SUBTAXGEXT  109 

Now  in  a  problem  of  limits  the  chord  of  an  infinitesimal  arc 
can  be  substituted  for  the  arc,  since  the  limit  of  their  ratio  is 
unity  (Art.  22  and  Cor.,  Art.  46) ;  so 

Ap  MQ  sin  MPQ. 

tCAs         chord  PQ         sinPJ/Q' 

,     .  ,/pAfl      u  chord  .VP      ..sin.VQP 

Aga.n,       B-^-  =  ft  chord  pQ  =  B  3"^^^^  i 

•••  S -««•-£■  (2) 

From  (1)  and  (2),  it  follows  that,  if  PZ  is  taken  as  ds, 
ds  =  PZ,     dp  =  HZ,     and     p  d0  =  #P. 

Drawing  OT  perpendicular  to  OP,  and  PA  and  (XV  perpen- 
dicular to  the  tangent  TP,  the  length  P77  is  the  polar  tan- 
gent; PA,  the  polar  normal;   OA,  the  pofar  subnormal;   and 
077,  the  pofa?*  subtangent. 
From  the  right-angled  HPZ 

ds2  =  dp2  +  p2dd-2;  (3) 

.    ,     pde         ,     dp     ,     .     pde  ,.. 

sm^  =  ¥'    cos^  =  5?    tan*  =  ^T  (4) 

Polar  subt.    =  OT  =  OP  tan  ^  =  p2  dd/dp.  (5) 

Polar  subn.   =  OA  =  OP  cot  ^  =  dp/dd.  (6) 


Polar  tan.      =  PT  =  VOP2  +  OT2  =  p  \J  1  -+ 


dfl2 
dp2 


Polar  norm.  =  AP  -  VOP2  +  OA2  =  y  p2  +  ^-  (8) 

p  =  ON  =  OP  sin  ^  =  p2  dd/ds 
P2 


Vp^  +  (dP/^)2'  (9) 

<t>  =  ^  +  d.  (10) 

Corollary.  —  If  PZ  represents  the  velocity  at  P  of  a  moving 


110  DIFFERENTIAL  CALCULUS 

point  (p,  0)  along  its  path,  PK  ( =  HZ)  and  PR  will  repre- 
sent its  component  velocities  at  P  along  the  radius  vector 
and  a  line  perpendicular  to  it. 

If  the  path  is  a   circle  with  center  at  0,  \J/  is  90°;  and 

•     ,        •    ™o      -.       pdd    ,         fAS        ds         dd 
sm^  =  sin  90 °  =  1  =  —-,  from  (4),  or  -=-  =  p-j->    .*.   v  =  rw; 

that  is,  the  linear  velocity  =  radius  times  angular  velocity. 
(See  Art.  72.) 

EXERCISE   X. 

1.  Find  the  subtangent,  subnormal,  tangent,  normal,  and  p  of  the 
spiral  of  Archimedes  p  =  ad.  _____ 

Ans.  subt.  =  p2/a;  subn.  =  a;  normal  =  Vp2  +  a2; 
tangent  =  p  Vl  +  p2/a2;  p  =  p2/(p2  +  a2)*- 

2.  In  the  spiral  of  Archimedes  show  that  tan  ^  =  0;  thence  find  the 
values  of  \p,  when  6  =  2-w  and  4 tt. 

Ans.  80°  57'  and  85°  27'. 

3.  Find  the  subtangent,  subnormal,  tangent,  and  normal  of  the 
logarithmic  spiral  p  =  ae.  

Ans.  subt.  =  p/loga;    subn.  =  ploga;  tan.  =  p  Vl  +  (log  a)-2; 

norm.  =  p  Vl  +  (log  a)2. 

4.  Show  why  the  logarithmic  spiral  is  called  the  equiangular  spiral, 
by  finding  that  \f/  is  constant. 

If  a  =  e,  $  =  ir/4,  subt.  =  subn.,  and  tan.  =  norm. 

5.  Find  the  subtangent,  subnormal,  and  p  of  the   Lemniscate  of 
Bernouilli  p2  =  a-  cos  2  6. 

Ans.  subt.  =  —  p3/a2sin2  0;     subn.  =  —  a2sin2  0/p; 
p  =  p3/Vp4  +  a4  sin2  2  6  =  Pz/a2. 

6.  In  the  circle  p  =  a  sin  6,  find  \p  and  <£. 

Ans.  ^  =  0,  and  0  =  20. 
The  angle  between  two  polar  curves  is  found  as  for  the  other  curves. 

7.  Find  the  angle  of  intersection  between  the  circle  p  -  2a  cos  0, 
and  the  cissoid  p  =  2  a  sin  0  tan  0. 

Ans.  arc  tan  2. 


CHAPTER  V. 
MAXIMA  AND   MINIMA.     INFLEXION  POINTS. 

78.  Maxima  and  Minima.  —  One  of  the  principal  uses 
of  derivatives  is  to  find  out  under  what  conditions  the  value 
of  the  function  differentiated  becomes  a  maximum  or  a 
minimum. 

This  is  often  very  important  in  engineering  questions, 
when  it  is  most  desirable  to  know  what  conditions  will  make 
the  cost  of  labor  and  material  a  minimum,  or  will  make 
efficiency  and  output  a  maximum. 

A  maximum  value  of  a  function  or  variable  is  defined  to 
be  a  value  greater  than  those  values  immediately  before  and 
after  it,  and  a  minimum  value  to  be  one  less  than  those 
immediately  before  and  after  it.  It  follows  that  the  function 
is  increasing  before,  and  decreasing  after  reaching  a  maxi- 
mum value;  while  it  is  decreasing  before,  and  increasing 
after  reaching  a  minimum  value. 

The  points  on  the  graph  of  y  =  f  (x)  at  which  the  function 
ceases  to  increase  and  begins  to  decrease,  or  ceases  to  de- 
crease and  begins  to  increase,  are  maxima  or  minima  points ; 
and  the  values  of  the  function  at  those  points  are  maxima 
or  minima  values. 

It  is  to  be  noted  that  a  maximum  value  is  not  necessarily 
the  greatest  value  the  function  can  have  nor  a  minimum  the 
least;  f  (a)  is  a  maximum  if  it  be  greater  than  any  other  value 
of/  (x)  near/  (a)  and  on  either  side  of  it;  and/  (a)  is  a  mini- 
mum if  it  be  less  than  any  other  value  of  /  (x)  near  /  (a)  and 
on  either  side  of  it. 

79.  The  Condition  for  a  Maximum  or  a  Minimum  Value. 
—  If  /  (x)  is  a  function  of  an  increasing  variable  x;  then 

111  - 


112  DIFFERENTIAL  CALCULUS 

for  /  (a)  to  be  a  maximum,  /  (x)  must  be  increasing  just 
before  /  (a)  and  therefore  /'  (x)  must  be  positive ;  on  the 
other  hand  /  (x)  must  be  decreasing  just  after  /  (a)  and 
therefore  /'  (x)  must  be  negative.  Hence,  as  x  increases 
through  the  value  a,  f  (x)  must  change  from  a  positive  to  a 
negative  value.  Conversely,  if  as  x  increases  through  the 
value  a,  f  (x)  changes  from  a  positive  to  a  negative  value, 
/  (a)  will  be  a  maximum  value  of  /  (x) . 

Hence  /  (a)  will  be  a  maximum  value  of  /  (x)  if,  and  only 
if,  /'  (x)  changes  from  a  positive  to  a  negative  value  as  x 
increases  through  the  value  a. 

In  the  same  way  it  may  be  seen  that/  (a)  will  be  a  minimum 
value  of  /  (x)  if,  and  only  if,  /'  (x)  changes  from  a  negative  to 
a  positive  value  as  x  increases  through  the  value  a. 

This  condition  has  been  called  the  fundamental  condition 
or  test. 

For  the  cases  of  most  frequent  occurrence;  when  f  (a)  is  a 
maximum  or  a  minimum,  f  (a)  =  0.  In  most  cases  it  is  a 
necessary  condition  for  a  maximum  or  a  minimum  value  of 
a  function  that  the  first  derivative  at  that  value  shall  be 
zero.  For  in  most  cases  the  first  derivative  /'  (x)  is  con- 
tinuous; and,  when  continuous,  it  changes  sign  by  passing 
through  the  value  zero  only.  But  if  /'  (x)  is  not  continuous, 
as  is  the  case  for  some  functions,  then  it  may  change  sign  by 
becoming  infinite  for  some  finite  value  of  x;  for  if/'  (x)  is 
a  fraction  whose  denominator  becomes  zero  for  some  finite 
value  of  x,  f  (x)  changes  sign  as  x  increases  through  that 

value.     For  example,  when  /'  (x)  =  5 ,  for  x  =  a  =  2, 

X         Zi 

f  (a)  =  - -  =  00  ;     here  /'  (x)    is   negative    before,    and 

positive  after  x  increases  through  the  value  2;  hence,  /  (2) 
is  a  minimum  according  to  the  fundamental  test.  Again, 
there  are  exceptional  non-algebraic  functions  for  which/'  (x), 
as  x  increases  through  some  finite  value  a,  changes  sign 


GRAPHICAL  ILLUSTRATION 


113 


without  becoming  either  zero  or  infinite.  (See  Note, 
Art.    80.) 

Excepting  such  rare  functions,  a  theorem  may  be  stated 
thus: 

For  all  algebraic  Junctions  any  value  of  x  which  7tiakes  f  (x) 
a  maximum  or  a  minimum  is  a  root  off  (x)  =  0  or  f  (x)  =  go  . 

The  converse  of  this  theorem  is  not  true ;  that  is,  any  root 
of  /'  (x)  —  0  or  /'  (x)  =  oo  does  not  necessarily  make  /  (x) 
either  a  maximum  or  a  minimum.  These  roots  are  called 
critical  values  of  x,  and  each  root  may  be  tested  by  rule. 

80.  Graphical  Illustration.  —  Let  P  ...  P3  ...  P?  be 
the  locus  of  y  =  f  (x) .  Then  /  (x)  will  be  represented  by  the 
ordinate  of  the  point  (x,  y),  and  /'  (x)  by  the  slope  of  the 
locus  at  the  point  (x,  y).  By  definition,  the  ordinates  MP, 
MzP2,  and  M4P4  represent  maxima  of  /  (x) ;  while  0,  MiPi, 
M3P3,  and  MbPb  represent  minima.     (Art.  78.) 


The  slope  /'  (x)  is  positive  immediately  before  a  maximum 
ordinate,  and  negative  immediately  after;  while  the  slope  is 
negative  immediately  before  a  minimum  ordinate  and  posi- 
tive after.  The  slope  /'  (x)  is  0  or  00  at  any  point  whose 
ordinate  /  (x)  is  either  a  maximum  or  a  minimum.  The 
slope  f  (x)  is  discontinuous  at  the  points  P4  and  P5,  where 
it  changes  sign  by  becoming  infinite  as  x  increases  through 
the  values  OMt  and  0M-0)  that  is,  /'  (x)  =  go  . 


114 


DIFFERENTIAL  CALCULUS 


The  slope  f  (x)  is  0  at  P6  and  oo  at  P7;  but  it  does  not 
change  sign  at  either  point,  and  neither  M6Pe  nor  M7P7  is  a 
maximum  or  a  minimum  ordinate;  it  does,  however,  change 
in  value  at  each  point,  P6  being  a  point  where  the  slope  f  (x)  is 
a  minimum  and  P7  one  where  it  is  a  maximum.  The  points 
P6  and  P7  are  inflexion  points,  at  which  the  curve  changes 
from  being  concave  downward  to  upward,  or  vice  versa. 

Note.  —  Points  such  as  P4  and  P5  occur  on  railroad  "Y's," 
and  such  points  where  branches  of  a  curve  end  tangent  to 

each  other  are  called  cusps. 
At  a  point  on  a  non-algebraic 
curve  where  branches  end  and 
are  not  tangent  to  each  other, 
called  a  shooting  point,  f  (x) 
may  change  abruptly  from  a 
positive  finite  value  to  a  nega- 
tive value,  or  vice  versa;  hence, 
/  (a)  would  be  a  maximum  or  a  minimum  without  /'  (x) 
becoming  either  zero  or  infinite.  The  supplementary  figure 
shows  a  shooting  point  at  which  /  (a)  is  a  minimum;  /'  (x) 
becoming  —  1  as  x  increases  to  a,  and  + 1  as  x  decreases  to 
the  same  value  a,  thus  changing  from  a  negative  to  a  positive 
value  as  x  increases  through  the  value  a. 

It  is  to  be  noted  that,  while  on  an  exceptional  curve  like 
the  one  shown  the  tangents  at  a  maximum  or  a  minimum 
point  may  have  various  directions,  on  any  algebraic  curve 
the  tangent  is  parallel  to  one  or  other  of  the  two  rectangular 
axes;  that  is,  the  tangent  at  a  maximum  or  a  minimum  point 
is  horizontal,  the  slope  being  continuous;  otherwise  it  is 
vertical ;  and  on  only  exceptional  non-algebraic  curves  will  it 
have  any  other  direction. 

It  may  be  noted  also,  as  in  the  graphical  illustration 
given,  that  maxima  and  minima  occur  alternately;  that  is, 
a  minimum  between  any  two  consecutive  maxima  and  vice 
versa.     It  may  be  seen  that  a  maximum  may  be  less  than 


RULE   FOR  APPLYING  FUNDAMENTAL  TEST        115 

some  minimum  not  consecutive,  since  by  definition  it  is 
necessarily  greater  than  those  values  only  immediately  before 
and  after  it.  It  may  be  seen  also  that  when  the  slope  is 
continuous  at  least  one  inflexion  point  must  occur  between 
a  maximum  and  a  minimum  point.  The  only  inflexion 
points  marked  on  the  curve  are  P6  and  P7,  occurring  where 
/'  (x)  =  0  and  oo ,  but  /'  (x)  may  have  any  value  at  an 
inflexion  point,  although  its  rate,  j"  (x),  must  change  sign 
there,  becoming  0  or  oo .  Hence,  at  any  inflexion  point,  a 
point  where  the  slope  f  (x)  is  a  maximum  or  a  minimum, 
5"  (x)  —  0  or  oo .  The  converse  is  not  true,  for  J"  (x)  may 
be  0  or  oo  at  other  points. 

81.  Rule  for  Applying  Fundamental  Test.  —  Let  a  be  a 
critical  value  given  by  either  f  (x)  =  0  or  f  (x)  =  oo ,  or,  in 
general,  any  value  of  x  to  be  tested,  and  Ax  a  small  positive 
number;  then: 

If  f  (a  ~  Ax)  is  positive  and  f  (a  +  Ax)  is  negative, 

f  (a)  is  a  maximum  off  (x) .  (Art.  79.) 

Iff  (a  ~~  Ax)  is  negative  andf  {a  +  Ax)  is  positive, 

f  (a)  is  a  minimum  of  f  (x).  (Art.  79.) 

If  f  (a  ~  Ax)  and  f  (a  +  Ax)  are  both  positive  or  both 
negative,  f  (a)  is  neither  a  mvaximum  nor  a  minimum  of 
fix). 

This  rule  is  general  and  is  valid  for  all  functions  that  are 
continuous  one-valued  functions,  which  comprise  all  those 
usually  encountered  in  this  connection. 

82.  While  the  rule  just  stated  applies  in  every  case; 
when  /'  (x),  as  well  as  /  (x),  is  continuous  and  therefore  the 
criticalyalues  of  x  are  roots  of  f  (x)  =  0,  a  rule  usually  easier 
to  apply  may  be  deduced  from  the  fundamental  test  or 
condition. 

Let  a  be  a  critical  value  of  x  given  by  /'  (x)  =0.  If  /  (a) 
is  a  maximum  value  of  /  (x) ,  /'  (x)  changes  from  a  positive 
to  a  negative  value  as  x  increases  through  a ;  therefore,  near 
a,  f  (x)  is  a  decreasing  function,  and  therefore,  its  derivative, 


116  DIFFERENTIAL  CALCULUS 

/"  (x),  must  be  negative  near  a.  But  if  /"  (a)  is  not  zero, 
then  near  a  the  sign  of  /"  (x)  is  that  of  J"  (a).  Hence 
f"  (a),  if  it  is  not  zero,  will  be  negative  when/  (a)  is  a  maxi- 
mum value  of  /  (x). 

In  the  same  way  it  is  seen  that/"  (a),  if  it  is  not  zero,  will 
be  positive  when  /  (a)  is  a  minimum  value  of  /  (x) . 

Conversely,  /  (a)  will  be  a  maximum  or  a  minimum  value 
of  /  (x)  according  as  /"  (a)  is  negative  or  positive. 

Hence  this  rule  for  determining  the  maxima  and  minima 
values  of/  (x)  when/  (x),  f  (x)  are  continuous: 

The  roots  of  the  equation  f  (x)  =  0  are,  in  most  cases,  the 
values  of  x  which  make  f  (x)  a  maximum  or  a  minimum. 

If  a  be  a  root  of  f  (x)  =  0;  then  f  (a)  will  be  a  maximum 
value  of  f  (x),  if  f"  (a)  is  negative,  but  a  minimum,  if  f"  (a)  is 
positive. 

83.  While  the  above  rule  is  all  that  is  needed  in  most 
cases,  it  does  not  provide  for  the  case  when  the  critical  value 
a  makes/"  (x)  become  zero.  When/"  (a)  =0,/  (a)  may  be 
either  a  maximum  or  a  minimum,  or  it  may  be  neither,  and 
the  point  on  the  graph  of/  (x)  may,  or  may  not,  be  a  point  of 
inflexion;  so  an  extension  of  the  rule  is  needed  to  provide  for 
cases  where  /"  (x)  and  the  succeeding  derivatives  may  in 
turn  become  zero  for  the  value,  a. 

If  no  derivative  is  found  that  does  not  become  zero  when 
a  is  substituted  for  x,  then  recourse  may  be  had  to  the  funda- 
mental test,  that  rule  applying  in  every  case.  But  if/'  (a), 
/"  (a),  .  .  .  ,  fn~l  (a)  all  are  found  to  be  zero,  and/n(a)  not 
zero;  then  the  following  rule,  inclusive  of  the  preceding, 
applies.  Let  a  be  a  critical  value  of  x  given  by  f  (x)  =  0, 
and  let  a  be  substituted  for  x  in  the  successive  derivatives 
of  f  (x). 

If  the  order  n  of  the  first  of  the  derivatives  that  is  not  zero  is 
an  even  integer,  f  (a)  will  be  a  maximum  or  a  minimum  off  (x) 
according  as  this  derivative  is  negative  or  positive. 

If  the  order  n  of  the  first  of  the  derivatives  that  is  not  zero  is 


RULE   FOR   DETERMINING   MAXIMA  AND   MINIMA    117 

an  odd  integer,/  (a)  will  be  neither  a  maximum  nor  a  minimum 
of  f  (x)  regardless  of  the  sign  of  this  derivative. 

Note.  —  This  conclusion  can  be  deduced  by  examining 
the  signs  of  the  derivatives  near  a;  thus,  as  follows: 

If  /'  (a)  and  /"  (a)  be  zero  but  /'"  (a)  not  zero;  since 
f"  (x),  the  rate  of  /"  0),  has  when  a:  is  a  a  value  not  zero; 
f"  (x),  the  rate  of  /'  (x),  is  then  increasing  or  decreasing 
according  as  /'"  (a)  is  positive  or  negative,  and,  since  it  is 
zero  when  x  is  o,  it  must  change  sign  as  x  increases  through  a; 
therefore,  /'  (x),  the  rate  of  /  (x),  must  be  either  decreasing 
before  and  increasing  after,  or  increasing  before  and  decreas- 
ing after  x  is  a,  and  so,  continuing  to  be  positive  or  negative 
according  as  f"  (a)  is  positive  or  negative,  does  not  change 
sign  as  x  increases  through  a;  hence/  (a)  is  neither  a  maxi- 
mum nor  a  minhnmu  of/  (x)  regardless  of  the  sign  of  /'"  (a). 

Now  if  /'"  (a)  also  is  zero  but  /IV  (a)  not  zero;  since  f™  (x), 
the  rate  of  /'"  (x),  has  when  x  is  a  a  value  not  zero,  /"'  (x), 
the  rate  of  f"  (x) ,  is  then  decreasing  or  increasing  according 
as  /IV  (x)  is  negative  or  positive  and,  since  it  is  zero  when  x 
is  a,  it  must  change  sign  as  x  increases  through  a;  therefore, 
/"  0),  the  rate  of/'  (x),  must  be  either  increasing  before  and 
decreasing  after,  or  decreasing  before  and  increasing  after  x 
is  a,  and  so,  continuing  negative  or  positive  according  as 
/JV  (a)  is  negative  or  positive,  does  not  change  sign  as  x 
increases  through  a;  /'  (x),  the  rate  of  /  (x),  must  then  be 
either  decreasing,  or  increasing  before  and  after  x  is  a,  and, 
as  it  is  zero  when  x  is  a,  it  changes  from  a  positive  to  a  nega- 
tive value,  or  from  a  negative  to  a  positive  value,  as  x  in- 
creases through  a,  according  as  /IV  (a)  is  negative  or  positive; 
hence  /  (a)  is  a  maximum  or  a  minimum  of  /  (x)  according 
as  /IV  (a),  the  first  of  the  derivatives  that  is  not  zero,  is  neg- 
ative or  positive. 

In  the  same  way  it  follows  that,  if  f*  (a)  is  the  first  of  the 
derivatives  that  is  not  zero,  /  (a)  is  neither  a  maximum  nor 
a  minimum  of  /  (x)  regardless  of  the  sign  of  f*  (a) ;  and  that, 


118 


DIFFERENTIAL  CALCULUS 


if /VI  (a)  is  the  first,  /  (a)  is  a  maximum  or  a  minimum  of/  (x) 
according  as  /VI  (a)  is  negative  or  positive ;  and  so  on  for  the 
succeeding  derivatives:  hence  the  inclusive  rule  given. 
(For  proof  by  Taylor's  Theorem,  see  Art.  218.) 


y=ffxh4x* 

ffa)=ffoJ=o,  neither  a  max.noraminoff(x) 

Y] 


y=f(x)=x* 
ffa)  =f(o)=o,  a  mm.  of  (x) 

r 


f"(x)=-12x2 


f(x)=-Z4x 


-2SL 
ffa)  ff/oJ=o,  neither  a  rruix.  nor  a  mfn ,  offfxj 


fJxJ-24 


V=f(x)  =  -x4 
ffa)  -SlpJ =o,a  max.  off(x) 


84.  Typical  Illustrations.  —  The  foregoing  deductions 
may  be  verified  by  the  graphs  of  the  successive  derivatives 
of  x3}  —Xs,  xA,  and  —  x4,  referred  to  the  same  axes  as  those  of 
the  graphs  of  the  functions.     The  usual  case  when  /'  (a)  is 


TYPICAL  ILLUSTRATIONS 


119 


zero  and  /"  (a)  not  zero,  is  well  illustrated  by  the  graphs  of 
the  function  sin  0  and  its  derivatives. 


f(e)~sinf) 


f(w<os  e 


f(d)=-sin6 


f  (0)  =  sin  0 ;  where  0  is  in  radians. 
fid)  =  cos0  =  0;     .'.     0  =  tt/2,  Itt,  .  .  .  ,  (tt/2  +  »x); 

.-.      /  (tt/2)  =  1  is  a  maximum  value  of  /  (x). 
ft  (|  tt)  =  -j- ;  _\f(%Tc)=-—  1  is  a  minimum  value  of/  (x) . 
^y^j=  -smd  =  0;     .'.      0  =  0,ir,  .  .  .,  fir; 

/.       (0,  0),  (r,  0),  .  .  •  are  points  of  inflexion. 

/"'(0)=  -  cos  0;/'"(o)  =  -; 

...     /'  (o)  =  1  is  a  maximum  slope, 
j///  (y)  _  _j_  .   ...     j'  (V)  =  -  1  is  a  minimum  slope. 

The  graphs  make  manifest  that  for  a  maximum  or  a  mini- 
mum value  of  the  function  the  first  derivative  passes  through 
zer^ibeing  +  before  and  -  after  for  a  max.,  and  -  before  and 


120  DIFFERENTIAL  CALCULUS 

+  after  for  a  min.;  and  that  hence  the  second  derivative  is 
—  for  a  max.  and  +  for  a  min. ;  also  that  at  an  inflexion  point 
the  second  derivative  is  zero.  The  graph  of  sin  6  makes 
manifest  at  once  the  maxima  and  minima  values  of  the 
function  and  the  value  of  the  angle  in  radians  that  makes  the 
function  a  maximum  or  a  minimum. 

The  graph  of  the  first  derivative  shows  that  the  first 
derivative  of  any  continuous  function  when  continuous  itself, 
changes  sign  by  passing  through  zero  only,  for  the  ordinate 
changes  sign  by  becoming  zero  only,  as  the  graph  crosses  the 
axis  of  0;  and  it  shows  that  in  passing  through  zero  the 
ordinate  changes  from  plus  to  minus,  or  from  minus  to  plus 
according  as  the  abscissa  of  the  point  of  crossing  corresponds 
to  a  maximum  or  a  minimum  ordinate  of  the  graph  of  the 
function ;  and  that  as  it  crosses  the  axis  the  ordinate  is  either 
decreasing  or  increasing.  The  graph  of  the  second  derivative 
shows  by  the  direction  or  sign  of  its  ordinate  at  or  near  the 
point  where  the  graph  of  the  first  derivative  crosses  the  axis 
whether  the  first  derivative  is  decreasing  or  increasing  as  it 
passes  through  zero  at  that  point,  the  sign  being  minus  or 
plus  according  as  the  first  derivative  is  decreasing  or  increas- 
ing; that  is,  according  as  the  abscissa  of  the  point  corre- 
sponds to  a  maximum  or  a  minimum  ordinate  of  the  graph 
of  the  function. 

85.  Inflexion  Points.  —  Where  the  slope  of  the  graph  of 
the  function  is  a  maximum  or  a  minimum  is  at  the  points 
where  the  ordinate  of  the  first  derivative  has  its  greatest 
positive  or  negative  value,  and  those  points  are  precisely 
where  the  ordinate  of  the  second  derivative  is  zero,  those 
values  of  6  that  make  the  slope  of  the  function  a  maximum 
or  a  minimum  being  those  that  make  its  derivative,  the 
second,  zero.  These  points  are  inflexion  points  on  the  graph 
of  the  function,  where  the  curve  changes  from  being  concave 
upward  to  downward,  or  downward  to  upward.  When  a 
curve  as  mn  is  concave  upward  its  slope  evidently  increases 


INFLEXION   POINTS 


121 


as  the  abscissa  of  the  generating  point  increases;  hence  its 
derivative,  the  second  (the  flexion),  is  positive.  When  a 
curve  as  st  is  concave  downward  its  slope  evidently  decreases 
as  the  abscissa  increases;  hence  the  second  derivative  is 
negative.  At  a  point  of  inflexion,  as  P  on  mt  or  sn,  the  tan- 
gent crosses  the  curve  at  the  point  of  contact,  and  on  opposite 


sides  the  curve  is  concave  in  opposite  directions,  therefore, 
the  second  derivative  has  opposite  signs.  Hence,  at  a  point 
of  inflexion  the  first  derivative,  the  slope  of  the  curve,  has  a 
maximum  or  a  minimum  value.  To  test  a  curve  for  points 
of  inflexion  is  to  test  its  slope  for  maxima  and  minima.  In 
the  case  of  roads  or  paths  that  change  direction  of  curvature 
in  a  horizontal  plane  the  inflexion  point  is  usually  called  the 
point  of  reverse  curvature,  and  when  the  curved  grade 
changes  direction  of  curvature  in  a  vertical  plane  the  inflexion 
point  is  where  the  grade  is  greatest  or  least  on  that  part  of 
the  road.  The  roots  of  the  second  derivative  =  0  or  x  are 
the  critical  values  to  be  tested  for  points  of  inflexion,  and 
the  sign  of  the  third  derivative,  when  the  critical  value  is 
given  by  the  second  derivative  =  0,  indicates  whether  the 
second  derivative,  the  flexion,  is  decreasing  or  increasing  as 
the  critical  value  is  passed,  the  sign  being  minus  or  plus 


122 


DIFFERENTIAL  CALCULUS 


according  as  the  second  derivative  is  decreasing  or  increas- 
ing; that  is,  according  as  at  the  critical  value  the  slope  is  a 
maximum  or  a  minimum,  or  as  the  curve  is  concave  upward 
before  and  downward  after  or  the  reverse.  These  conclu- 
sions are  verified  by  the  graphs  of  sin  0  and  its  successive 
derivatives. 

When  the  critical  value  is  given  by  the  second  derivative 
=  oo,  or,  in  general,  when  any  value  is  to  be  tested,  the 
fundamental  test  may  be  applied  to  determine  the  sign  of 
the  second  derivative  before  and  after  the  value  to  be  tested, 
and  thus  to  determine  the  concavity  and  existence  or  non- 
existence of  inflexion. 

It  is  evident  that  at  a  maximum  point  on  a  curve  the 
curve  is  concave  downward  both  sides  of  the  horizontal 
tangent  point  and  concave  upward  both  sides  of  a  vertical 
tangent  point,  while  the  reverse  is  the  case  at  a  minimum 
point. 


Tangents  drawn  at  successive  points  on  a  curve  that  has 
maxima  and  minima  points  show  that  as  the  abscissa  of  the 
moving  point  increases  the  tangent  turns  clockwise  through 
zero  angle  at  a  maximum  point  until  an  inflexion  point  is 
reached  when  it  f urns  in  the  opposite  direction  through  zero 
angle  again  at  a  minimum  point;  and  then  it  may  without 
any  inflexion  point  turn  through  a  right  angle  at  a  maximum 
point;  continuing  to  turn  anti-clockwise  until  possibly  at  an 
inflexion  point  it  turns  back  clockwise  through  a  minimum 
point  when  the  angle  is  a  right  angle  again,  becoming  less  as 


POLAR   CURVES 


123 


the  tangent  continues  to  turn  clockwise.  The  points  on  a 
graph  at  which  the  ordinate  ceases  to  increase  and  begins  to 
decrease,  or  else  ceases  to  decrease  and  begins  to  increase  are 
sometimes  called  turning  points  of  the  graph,  and  the  corre- 
sponding values  of  the  function  turning  values.  The  turning 
values  are  evidently  maxima  and  nn'nima  values  and  the 
turning  points  maxima  and  minima  points.  While  the  tan- 
gent at  an  inflexion  point  turns  in  opposite  directions,  the 
curve  is  either  rising  or  falling  on  both  sides  of  the  point;  but 
at  a  turning  point  the  curve  is  rising  at  a  maximum  and  then 
falling,  or  falling  at  a  minimum  and  then  rising.  These 
considerations  make  it  obvious  that  at  a  maximum,  the 
angle  made  by  the  tangent  decreasing,  its  rate,  the  second 
derivative  of  the  function,  is  negative,  and  that  at  a  mini- 
mum, the  angle  increasing,  the  second  derivative  is  positive. 


86.  Polar  Curves.  —  A  polar  curve  is  concave  or  convex 
to  the  pole  at  a  point,  according  as  the  tangent  to  the  curve 
at  the  point  does  not,  or  does  lie  on  the  same  side  of  the 
curve  as  the  pole.  It  may  be  seen  from  the  figure  that 
when  a  polar  curve,  as  mn,  is  concave  to  the  pole,  p  or  ON 
increases  as  p  increases;  hence,  the  rate  of  change  of  p  with 

respect  to  p,  -J-  is  positive. 
dp 

When  a  curve,  as  st,  is  convex  to  the  pole,  p  decreases  as 

p  increases;  hence,  -j-  is  negative. 
dp 


124  DIFFERENTIAL  CALCULUS 

It  follows  that  a  polar  curve  is  concave  or  convex  to  the  pole 

at  a  point  according  as  -j-  is  positive  or  negative. 

dp  dp 

At  a  point  of  inflexion  on  a  polar  curve,  as  P  on  mt,  -j- 

changes  sign,  and  therefore  p  is  a  maximum  or  a  minimum; 
and  conversely.  Hence  to  test  a  polar  curve  for  points  of 
inflexion,  p  is  tested  for  maxima  and  minima. 

Example.  —  Examine  the  Lituus  for  points  of  inflexion. 
_  p2  2a2p 

Here  P  -  Vp2  +  {dp/def  ~  V4a4  +  p4 ' 

fr      2a»(4«W)  .  aV2 

dp         (4  a4  +  p4)* 
Hence,  p  =  aV2  makes  p  a  maximum;  and  (0V2,  §)  is  a 
point  of  inflexion. 

The  spiral  p  =  ae  has  no  point  of  inflexion,  since  -7-  is 

always  positive. 

87.  Auxiliary  Theorems.  —  By  use  of  the  following 
theorems,  which  are  obvious,  the  solutions  of  problems  in 
maxima  and  minima  are  often  simplified : 

(i)  Any  value  of  x  which  makes  c  +  /  (x)  a  maximum  or  a 
minimum  makes  /  (x)  a  maximum  or  a  minimum;  and 
conversely. 

(ii)  Any  value  of  x  which  makes  c  •/  (x),  c  being  positive, 
a  maximum  or  a  minimum  makes  /  (x)  a  maximum  or  a 
minimum;  and  conversely.  If  c  is  negative  and  /  (a)  is  a 
maximum,  C'f  (a)  is  a  minimum. 

(hi)  Any  value  of  x  which  makes  f(x)  positive,  and  a 
maximum  or  a  minimum,  makes  [f  (x)]n  a  maximum  or  a 
minimum,  n  being  any  positive  whole  number. 

(iv)  Since  f  (x)  and  \ogf(x)  increase  and  decrease  to- 
gether, any  value  which  makes  /  (x)  a  maximum  or  a  mini- 
mum makes  log/  (x)  a  maximum  or  a  minimum;  and  con- 
versely. 


EXERCISE  XI  125 

(v)  Since  when  /  (x)  increases  its  reciprocal  decreases,  any 
value  of  x  which  makes  /  (x)  a  maximum  or  a  minimum 
makes  its  reciprocal  a  minimmn  or  a  maximum. 

EXERCISE   XI. 

Examine  /  (x)  for  maxima  and  minima  when: 

1.  /(x)=x3-5.'c4  +  5x3-  1. 

fix)  =  5xi-20x?  +  lox°-  =  5x2(x2-4x  +  3) 

=  5x2(x-l)(x-3)  =  0;     .*.     x  =  0,1,3. 
/"  (x)  =  20  x3  -  60  x2  +  30  x)  /'"  (x)  =  60 x2  -  120  x  +  30. 
/"(0)  =0,  /'"(0)  =30;     .-.     /(0)  =  -1 
is  neither  a  max.  nor  a  min. 

/"  (1)  =  20  -  60  +  30  =  -10;      .".     /(I)  =  0  is  a  max. 
/"  (3)  =  540  -  540  +  90  =  90;      .*.     /  (3)  =  -28  is  a  min. 
By  plotting  the  graph  of/  (x)  these  results  may  be  verified. 

2.  f  (x)  =  x3  -  3  x2  +  3  x  +  7. 

fix)  =  3  x2  -  6  x  +  3  =  3  (x2  -  2  x  +  1) 

=  3  (x  -  l)2  =  0;     .'.     x  =  1, 1. 
/"  (x)  =  6x  -  6  =  6  (x  -  1);  /'"  (x)  =  6. 
/"  (1)  =  0;/'"  (1)  =  6;     /.     /(I)  =8  is  neither  a  max.  nor  a  min. 

3.  fix)  =3x*  -4x3  +  1. 

f'(x)  =  12 x3  -  12 x2  =  12x2(x  -  1)  =0;     .*.     x  =  0,  L 
/"(x)  =36x2  -24x  =  ^x^x^);/'"  (x)  =  72x  -  24. 
/"(0)  =0;/'"(0)  =  -24;      .-.     /(0)  =  1 
is  neither  a  max.  nor  a  min. 

/"(I)  =  12;      /.     /(I)  =0  is  a  min. 

4.  /(x)  =  3  j?  -  125x2  +  2160  x. 

f  (x)  =  15  x4  -  375  x2  +  2160  =  15  (x4  -  25  x2  4- 144)  =  0; 

.'.     x  =  ±3,  ±4. 
/"  (x)  =  15  (4  x3  -  50  x) ;  /"  (3)  is  neg. ;      /.     /  (3)  is  max. 
/"  (4)  is  positive;    .*.  /  (4)  is  min.;  /"  (-3)  is  positive;    .*.  /  (-3) 
is  min.;  /"  (  —  4)  is  negative;     /.     /  (-4)  is  max. 

5.  /(x)  =x3-3x2  +  6x  +  7. 

f  (x)=3x2-6x  +  6  =  3  (x2-  2x4-2)  =  0;     .*.    x  =  1  ±  V^i. 
Hence  no  real  value  of  x  makes  /  (x)  a  max.  or  a  min. 

6.  Examine  c  +  V4  a2x2  —  2  ax3  for  max.  and  min.. 
Let     / (x)  =2  ax2  -  x3  (by  Art.  87). 

f  (x)  =  4  ax  -  3  x2  =  4  (4  a  -  3  x)  =  0;     .'.     x  =  4/3  a,  0. 
/"  (x)  =  4  a  -  6  x;  /"  (0)  =  4  a;      .-.     /  (0)  =  c  is  a  min. 
/"(4/3a)  =  4a -8a  =  -4a; 

.'.     /  (4/3  a)  =  c  +  8  a2  V3/9  is  a  max. 


126  DIFFERENTIAL  CALCULUS 

7.  y  =  a  —  3  (x  —  c)*}  and  y  =  a  -  b  (x  —  c)K 

~y~  = 7  =  oo  ;     .*.     x  =  c,  and  it  can  be  seen  that  -^ 

dx  3  (x  -  c)1  dx 

changes  from  +  to  —  when  x  passes  through  the  value  c,  hence/  (c)  =  a 

is  a  max. 

When/(x)=a-6(x-c)i;  dJL=  _— ri  =  «>;      »     x  =  c; 

ax  3(x  — c)3 

dy 
here  it  can  be  seen  that  /'  (x)  or  j~  does  not  change  sign  as  x  passes 

through  c,  and,  therefore,  the  function  has  neither  max.  nor  min. 

8.  Examine  (x  -  l)4  (x  +  2)3  for  max.  and  min. 

f(x)  =  (3-l)"(x  +  2)*(7s  +  5)=0;     ...     x  =  1,  -2,  -f. 

/'  (1  -  Ax)  is  -,  /'  (1  +  Ax)  is  +  ;      .*.    /  (1)  =  0  is  a  min. 
/'  (-f  -  Ax)  is  +,  /'(-$+  Ax)  is  -  ;    .\  /(-f)  is  a  max. 
/'  (-2  -  Ax)  and  f  (-2  +  Ax)  are  both  +;   hence  /  (-2)  is 
neither  max.  nor  min. 

(q  x^ 

9.  Examine ^r~  for  max.  and  min. 

a  —  2x 

/'  (x)  =  (a  -  x)2  (4  x  -  a) /(a  -  2  x)2;  /'  (x)  =  0  gives  x  =  a,  a/4; 

/'  (x)  =  oo  gives  (a  —  2  x)2  =  0,  or  x  =  a/2. 

/'  (a/4)  changes  from  —  to  +  as  x  passes  through  a/4;  .'.  /(a/4) 
is  min.     When  x  =  a,  or  a/2,  /'(x)  does  not  change  sign;    .'.   /(a)  and 
/  (a/2)  are  neither  max.  nor  min. 
x2  _  7  x  _i_  6 

10.  Examine rr; —  for  maxima  and  minima. 

x  —  10 

Ans.  /  (4)  is  a  max.;  /  (16)  is  a  min. 
(2  _i_  2)3 

11.  Examine  ; =rs  for  maxima  and  minima. 

(x  -  3)2 

Ans.  /  (3)  is  a  max;  /  (13)  is  a  min. 

12.  When/  (x)  =  (x  -_1)  (x  -  2)  (x  -  3),  /  (2  -  I/V3)  .  §  V3, 
is  a  max.,  and/  (2  +  I/V3)  =  -§  V3  is  a  min. 

13.  Show  that  the  maximum  value  of  sin  0  +  cos  6  is  V2. 

14.  Show  that  the  maximum  value  of  a  sin  0  +  b  cos  0  is  Va2  -f-  o2. 

15.  Show  that  e  is  a  minimum  of  x/log  x. 

16.  Show  that  1/ne  is  a  maximum  of  log  x/xn. 

17.  Show  that  e1/c  is  a  maximum  of  x1  x. 

18.  Show  that  1  is  a  maximum  of  2  tan  0  —  tan2  0. 

19.  Find  the  maximum  value  of  tan-1  x  —  tan-1  x/4,  the  angles 

being  taken  in  the  first  quadrant. 

Ans.  tan-1 1. 

20.  Show  that  2  is  a  maximum  ordinate  and  —26  is  a  minimum 
ordinate  of  the  curve  y  =  x6—-5xi  +  5x3  +  l. 


EXERCISE   XI 


127 


PROBLEMS   IN   MAXIMA  AND   MINIMA. 

1.    Find  the  maximum  rectangle  that  can  be  inscribed  in  a  circle  of 
radius  a.     Let  2  i  =  base  and  2  y  =  altitude;   then 
area  A  =  4  xy  =  4  x  Va2  —  x2. 


Take       /  (x)  =  x2  (a2  -  x2)  =  a2x2  -  x4  [by  Art.  87]; 

f  (x)  =  2  a2x  -  4  x3  =  2  x  (a2  -  2  x2)  =  0;    .-.    x  =  0,  a/V2] 
f"(x)  =  2  a2  -  12  x2;  /"  (0)  =  2  a2;      .'.     /  (0)  =  0  is  min.     ' 
f"  (a/y/2)  =  2a2  -  6a2  =  -4a2;     .'.     /(0/V2)  =  a4/4. 
.*.     A  =  4  VaV4  =  2  a2 

is  the  area  of  the  maximum  rectangle,  which  is  a  square. 
Note.  —  By  Geometry  without  the  Calculus  method: 

A  =  2  ay  =  2aVa— 'i5]^  =  2  a2, 

since  the  radical  quantity  is  evidently  greatest  when  x  =  0. 


2.  The  strength  of  a  beam  of  rectangular  cross  section  varies  as  the 
breadth  b  and  as  d2,  the  square  of  the  depth.  Find  the  dimensions  of 
the  section  of  the  strongest  beam  that  can  be  cut  from  a  cylindrical 


128 


DIFFERENTIAL  CALCULUS 


log  whose  diameter  is 

2  a. 

Strength  oc  6d2;    /. 

strength  =  kbd2 

A;  is  a  constant; 

let 

fff>)~ 

b  (4  a2 

-62) 

=  4a26 

-¥; 

r— 

f  (&)  = 

4a2- 

362  = 

=  0;     /. 

b  = 

2a 

V3' 

d  =  yi(2°)- 

f(b)  = 

3(< 

:a2- 

4a2\ 
3  J 

2a 

|(4i 

3V3 

where 


Hence,  the  rectangle  may  be  laid  off  on  the  end  of  the  log  by  drawing 
a  diameter  and  dividing  it  into  three  equal  parts;  from  the  points  of 
division  drawing  perpendiculars  in  opposite  directions  to  the  circum- 
ference and  joining  the  points  of  intersection  wfth  the  ends  of  the 
diameter,  as  in  the  figure.  The  strength  of  the  beam  is  about  0.65  of 
that  of  the  log,  but  it  is  the  strongest  beam  of  rectangular  section. 

3.  The  stiffness  of  a  rectangular  beam  varies  as  the  breadth  b  and  as 
d3,  the  cube  of  the  depth.  Find  the  dimensions  of  the  stiffest  beam  that 
can  be  cut  from  the  log. 


\fc-  v- 


Stiffness  <*  bd3;  ,%  stiffness  =  kbd3  (k  constant);  let  stiffness 
=  b  (4  a2  -  6«)3.    Take 

f(b)  =  4a%*  -  6§;  f  (b)  =  f  (a26""  -  6s)  »  0; 
.-.     62  =  a2     or     b  =  a)     :.     d  =  (4  a2  -  a2)*  =  a  V3. 

To  draw  the  rectangle,  lay  off  from  ends  of  a  diameter  chords  at  angles 
of  30°  with  diameter  and  join^nds_of_chords  with  ends  of  diameter. 

4.  A  square  piece  of  pasteboardis!^o~T5e~niaave  into  a  bcrc~by  cutting 
out  a  square  at  each  corner.  Find  the  side  of  the  square  cut  out,  so 
that  the  remainder  of  the  sheet  will  form  a  vessel  of  maximum  capacity. 
Let  a  be  side  of  square  sheet  and  x  side  of  square  cut  out;  then  ■ 


EXERCISE  XI 


129 


fix)  =  x(a-2x)2. 
f  (x)  =  -  4  x  (a  -  2  x)  +  (a  -  2  x)2  =  (a  -  6  x)  (a  -  2  z)  =  0; 

.'.     a  =  1/6  a,  1/2  a. 

/(l/6a)  =  1/6  a.  (a  -  1/3  a)2  =  2/27  a3,  maximum  capacity. 
/  (1/2  a)  =  1/2  a  (a  —  a)  =0,  minimum. 


cl-Zjc 

tZ 

X 

6.  A  rectangular  sheet  of  tin  15"  X  8"  has  a  square  cut  out  at  each 
corner.  Find  the  side  of  the  square  so  that  the  remainder  of  the  sheet 
may  form  a  box  of  maximum  contents. 

Arts.  If". 

6.   A  channel  rectangular  in  section,  carrying  a  given  volume  of 
water,  is  to  be  so  proportioned  as  to  have  a  mini- 
mum wetted  perimeter.     Find  the  proportions  of 
the  channel. 

Let  x  be  the  width  of  the  bottom,  and  y  the 
height  of  the  water  surface.  Since  the  given 
volume  is  proportional  to  the  cross  section, 

xy  =  V,   where   V  is  constant. 

2V 

p  *=  x  +  2y  =  x  +  -—,  from  (1); 


A 

— -— 

y 

—  — 

—  —  — . — 

</^ 

~  _zr^  rrv  r^_~  ~ 

< X > 

(1) 


that 


dx  x2         ' 


2V 


V2~V  =  V2 


2  xy,    or    x  =  2  y. 


xy] 


To  show  that  this  makes  p  a  minimum;  note  that  for  x2  <  2  V,  •£■  is 


dx 


dp 


negative,  and  for  x2  >  2  V,  j-   is  positive,  therefore  for  x2  =  2  V,  or 
x  =  2  y,  p  is  a  minimum. 


130  DIFFERENTIAL  CALCULUS 

7.   Find  the  dimensions  of  a  conical  tent  that  for  a  given  volume  will 
require  the  least  cloth. 

y  =  i  Trr2h',   .'.  h  =  S  V/irr2,  where  r  is  radius  of  base  and  h  altitude.     (1) 

S  =  Trr  W2  +  h*  =  irr  (r2  +  9  V2/*2^)'  =  (tt2/-4  +  9  F2/r2)*, 
$  denoting  lateral  surface;  let 

/  (r)  ^  ^r4  +  9  F2A2  (by  Art.  87), 


18  V2      n  V2    "/6J. 

V      7T 


/'(r)  =4ttV3--^  =0, 


/"  (r)  =  127r2r2  +  54  F2//-4,  positive  for  any  r; 
hence  r  =  -=-  y  —  makes  S  a  minimum. 

From  (1),    A  =  V  —  =  r  V^;  and  slant  height  =  r  V3. 

f       7T 


8.  From  a  given  circular  sheet  of  tin,  find  the  sector  to  be  cut  out  so 
that  the  remainder  may  form  a  conical  vessel  of  maximum  capacity. 

Ans.  Angle  of  sector  =  (l-  Vf)  2tt  =  66°  14'. 

9.  The  work  of  propelling  a  steamer  through  the  water  varies  as  the 
cube  of  her  speed;  show  that  her  most  economical  rate  per  hour  against 
a  current  running  n  miles  per  hour  is  3  n/2  miles  per  hour. 

Let  v  =  speed  of  the  steamer  in  miles  per  hour; 

then  cv3  =  work  per  hour,  c  being  a  constant; 

and  v  —  n  =  the  actual  distance  advanced  per  hour. 

Hence,         cvz/v  —  n  =  the  work  per  mile  of  actual  advance. 

Find  the  most  economical  speed  against  a  current  of  4  miles  per  hour. 

10.  The  cost  of  fuel  consumed  by  a  steamer  varies  as  the  cube  of  her 
speed,  and  is  $25.00  per  hour  when  the  speed  is  10  miles  per  hour.  The 
other  expenses  are  $100  per  hour.  Find  the  most  economical  speed. 
Let  C  =  cost  per  hour  for  fuel  at  speed  of  v  miles  per  hour; 


EXERCISE  XI  131 

then  C  :  $25  =  f3  :  (10)3;     .*.     C  =  $25  yVQO)3; 

F,  s       $25  r3     d  ,  SlOOd         .         .  .     ,.  .  ,  d  .   , 

f  d-)  =  — — —  •  -  -4 :    where  a  is  distance  and  -  is  hours. 

*  (10)3     v  v  v 

.,..         50      ,      100  tf      n. 

t-3  =  2000,     or     v  =  ^2000  =  12.6  miles  per  hr. 
/  (12.6)  =  $12 d  (approximately); 

hence  cost  for  one  hour  about  SI 50;  cost  for  running  10  miles  at  12.6 
miles  per  hour  about  SI 20,  while  the  cost  for  running  the  10  miles  at 
10  miles  per  hour  is  SI 25,  and  at  15  miles  per  hour  the  cost  for  running 
10  miles  is  about  S123. 

11.  The  amount  of  fuel  consumed  by  a  steamer  varies  as  the  cube  of 
her  speed.  When  her  speed  is  15  miles  per  hour  she  burns  4|  tons  of 
coal  per  hour  at  S4.00  per  ton.  The  other  expenses  are  S12.00  per  hour. 
Find  her  most  economical  speed  and  the  minimum  cost  of  a  voyage  of 
2080  miles.  Ans.  10.4  mi.  per  hr.;  S3600. 

12.  A  vessel  is  anchored  3  miles  off  shore.  Opposite  a  point  5  miles 
farther  along  the  shore,  another  vessel  is  anchored  9  miles  from  the 
shore.  A  boat  from  the  first  vessel  is  to  land  a  passenger  on  the  shore 
and  then  proceed  to  the  other  vessel.  Find  the  shortest  course  of  the 
boat. 

Let  hi  be  the  distance  the  boat  goes  from  first  vessel  to  shore  and  h2 
the  distance  from  the  shore  to  the  other  vessel;  then 

/  (x)  =h+h  =  (32  +  x*)*  +  [9^  +  (5  -  z)]*; 

ft  f    \  %  «3  % r\ . 

J  W  "  (9~+^  "  (81  +  (5  -  x¥)?  "    ' 
whence  x  =  ±  f ;     h  +  h2  =  -^  +  V  =  13  miles. 

13.  Find  the  number  of  equal  parts  into  which  a  given  number  n 
must  be  divided  that  their  continued  product  may  be  a  maximum. 
Let 

/»-©■'  ™=(i)H-lH 

:.     log  -  =  1 ;      -  =  e,     and    x  =  - , 
&  x         '      x  e 

hence  the  number  of  parts  is  n/e,  and  each  part  is  e. 


132  DIFFERENTIAL  CALCULUS 

DETERMINATION   OF  POINTS   OF  INFLEXION. 

1.  Examine  y  =  x3  —  3  x2  —  9  x  +  9;  for  points  of  inflexion. 

P  =  3x2-6z-9, 
dx 

pi  =  6x  -  6  =  6  (x  -  1)  =  0,      .\     x  =  1, 

is  abscissa  of  an  inflexion  point.     The  point  is  (1,  —2),  to  the  right  of 
which  the  curve  is  concave  upward. 

2.  Examine  x3  —  3  bx2  +  a?y  =  0,  for  points  of  inflexion. 

Ans.  (b,  2  b3/a?)  is  a  point  of  inflexion,  or  of  maximum  slope, 
to  the  right  of  which  the  curve  is  concave  downward. 

3.  Examine  y  =  c  sin  x  for  points  of  inflexion. 

Ans.  (0,  0),  (±7r,  0),  (±2tt,  0)  .  .  .  . 

8  a3 

4.  Examine  the  Witch  of  Agnesi,  y  =    .    ,    . — r,  for  inflexion  points. 

x2  +  4  a2 

Ans.   (±2a/\/3,  3  a/2);  concave  downward  between  these  points, 
concave  upwards  outside  of  them. 

Find  the  points  of  inflexion  of  the  following  curves: 

5.  (:r/a)2  +  (y/b)i  =  1.  Ans.  x  =  ±a/V2. 

6.  y  =  (x2  +  x)  e~x.  Ans.  x  =  0  and  x  =  3. 

tj  -ax  '   bx  a  2  (log  a   —  1°S  &) 

7.  y  =  e  ax  —  e~bx.  Ans.  ,  • 

a  —  b 

8.  ?/  =  z3/a2  +  x2. 

Ans.  (0,  0),  (a  V3,  3  a  Vf),  (-a  V3,  -3  a  VJ). 


CHAPTER  VI. 


CURVATURE.     EVOLUTES. 


88.   Curvature. 


The  flexion  (Art.   13),  b 


dm      d?y 
dx  =  dx2} 

being  the  rate  of  change  of  the  tangent  of  the  angle  made 
with  the  z-axis  by  the  tangent  to  a  curve,  is  one  measure  of 
the  bending  of  the  curve  at  the  point  of  tangency.  This 
measure,  however,  is  dependent  upon  the  position  of  the 
axes  and  would  change  if  the  axes  were  rotated. 

There  is  a  measure  of  the  bending  called  the  curvature, 
which  does  not  depend  upon  the  choice  of  the  axes,  as  it  is 
expressed  in  terms  that  are  the  same  after  the  axes  are 
rotated,  or  even  before  any  axes  are  drawn.     The  curvature 

is  denoted  by  K  =  — ,  the  rate  of  change  of  the  angle  of 

inclination    <f>  =  tan-1  m,    with 
respect  to  the  length  of  arc  s. 

Thus,  let  P  and  Pi  be  two 
points  on  a  plane  curve,  <f>  and 
</>  +  A$  the  angles  which  the 
tangents  at  P  and  Pi  make  with 
the  x-axis,  s  the  arc  AP  meas- 
ured from  some  fixed  point  A 
on  the  curve  up  to  P,  and  As 

the  arc  PPi.  The  angle  <f>  is  in  radians,  and  A<f>  is  evidently 
the  angle  between  the  two  tangents. 

The  angle  A<£  is  the  total  curvature  of  the  arc  As,  as  it  is  a 
measure  of  the  deviation  from  a  straight  line  of  that  portion 
of  the  curve  between  the  points  P  and  Pi.     The  sharper  the 

133 


134  DIFFERENTIAL  CALCULUS 

bending  of  the  curve  between  the  two  points  the  greater  is 
A0  for  equal  values  of  As. 

The  average  curvature  of  the  arc  As  is  denned  as  -r— ,  and 

is,  therefore,  the  average  change  per  unit  length  of  arc  in 
the  inclination  of  the  tangent  line. 

The  limit  of  -r— ,  when  Pi  approaches  P  as  its  limiting 

position,  is  called  the  curvature  of  the  curve  at  P;   that  is, 

the  curvature  at  a  point  on  a  curve  is  K  =  lim  -r—  =  -=-  • 

As=o  As      as 

Otherwise,  by  rates,  the  curvature  of  any  curve,  as  APPi 
at  any  point,  as  P,  is  the  s-rate  at  which  the  curve  bends  at 
P,  or  the  s-rate  at  which  the  tangent  revolves,  where  s  de- 
notes the  length  of  the  variable  arc  AP.  If  4>  denotes  (in 
radians)  the  variable  angle  XTP  as  P  moves  along  the  curve 
APPi,  then,  evidently,  the  curvature  of  APPi  at  P  equals 

the  s-rate  of  </> ;  that  is,  K  =  ~~- 

as 


89.   Curvature  of  a  Circle.  —  For  a  circle  of  radius  R 
As  =  PA0  and  therefore, 


A0_  1      d$_  l_ 

As  ~  R]    ds  ~  R' 

since  the  ratio  of  the  increments  is  con- 
stant; that  is,  the  average  curvature  of 
any  arc  of  a  circle  is  equal  to  the  curva- 
ture at  any  point  of  that  circle.  In 
other  words,  a  circle  is  a  curve  of  constant  curvature  and  its 
curvature  is  equal  to  the  reciprocal  of  its  radius;  that  is, 
the  curvature  of  a  circle  equals  1/R  radians  to  a  unit  of 
arc. 

For  example,  if  R  =  2,  the  circle  bends  uniformly  at  the 


CIRCLE,   RADIUS,   AND  CENTER  OF  CURVATURE      135 

If  R  =  J,  the  curvature  of  the  circle  is  2  radians  per  unit 
of  arc. 

If  R  =  1,  for  the  circle  of  unit  radius  the  curvature  is 
evidently  a  radian  per  unit  of  arc. 

90.  Circle,  Radius,  and  Center  of  Curvature.  —  The 
curvature  of  any  curve  except  the  circle  varies  from  one 
point  to  another.  A  circle  tangent  to  a  curve  and  having 
the  same  curvature  as  the  curve  at  the  point  of  contact, 
therefore,  having  a  radius  equal  to  the  reciprocal  of  the 
curvature  at  that  point,  is  called  the  circle  of  curvature  at 
that  point;  its  radius  is  called  the  radius  of  curvature;  and 
its  center,  the  center  of  curvature. 

If  R  denotes  the  radius  of  the  circle  of  curvature  at  any 

point  of  a  curve,  then,  since  the  curvature  of  the  curve  is  -=- 
and  equals  the  curvature  of  the  circle,  it  follows  that, 

d<f>       1  ,     0      ds 

ck=W    and    B  =  d* 

If  at  P  (Art.  72,  figure)  the  direction  of  the  path  of  (x,  y) 
became  constant,  (x,  y)  would  trace  the  tangent  at  P,  and  ds 
might  be  represented  by  a  length  on  the  tangent ;  while  if  at 
P  the  change  of  direction  of  the  path  became  uniform  with 
respect  to  s,  (x,  y)  would  trace  the  circle  of  curvature  at  P, 
and  ds  would  represent  an  arc  of  the  circle,  since  it  equals 
R  d<j),  where  d<f>  =  l<f>  would  be  the  constant  change  of  angle 
at  center  of  the  circle  of  curvature. 

Thus  it  is  that  the  curvature  is  uniform  when,  as  the 
moving  point  passes  over  equal  arcs,  the  tangent  turns 
through  equal  angles;  or  conversely;  and,  as  this  is  the  case 
with  the  circle  only,  it  is  the  only  curve  of  uniform  curvature. 

For  any  curve  the  measure  of  the  curvature  at  a  point  is 
the  limit  of  the  ratio  between  the  angle  described  by  the 
tangent  and  the  arc  described  by  the  point  of  contact,  as 

that  arc  approaches  zero;    and  this  limit  -=-  equals  the 


136 


DIFFERENTIAL  CALCULUS 


reciprocal  of  the  radius  of  curvature  at  the  point;    hence, 

ds 
the  radius  of  curvature  is  -7-  and  equals  the  reciprocal  of  the 

curvature.  The  figure  shows  the  circle  of  curvature  for  the 
point  P  (x,  y)  of  the  ellipse;  C  is  the  center  of  curvature, 
and  CP  the  radius  of  curvature.     It  is  to  be  noticed  that  the 


circle  of  curvature  crosses  the  ellipse  at  P,  and  this  must  be 
so;  for  at  P  the  circle  and  ellipse  have  the  same  curvature, 
but  towards  A  the  curvature  of  the  ellipse  increases,  while 
that  of  the  circle  remains  the  same,  being  constant.  Hence 
on  the  side  of  P  towards  the  vertex  A  the  circle  is  outside  of 
the  ellipse.  From  P  towards  B  the  curvature  of  the  ellipse 
decreases,  and,  therefore,  on  the  side  of  P  towards  the  vertex 
B  the  circle  is  inside  of  the  ellipse. 

So,  in  general,  the  circle  of  curvature  crosses  the  curve  at  the 
point  of  contact. 

The  only  exceptions  to  this  rule  are  at  points  of  maximum 
and  minimum  curvature,  as  the  vertices  of  the  ellipse.  From 
A  along  the  curve  in  either  direction,  the  curvature  of  the 
ellipse  decreases;  hence  the  circle  of  curvature  at  A  lies 
entirely  within  the  ellipse.  From  B  the  curvature  of  the 
ellipse  increases  in  each  direction  and  so  the  circle  of  curva- 
ture at  B  lies  entirely  without  the  ellipse. 


RADIUS  OF  CURVATURE  IN  COORDINATES       137 

91.  Radius  of  Curvature  in  Rectangular  Coordinates.  — 

Since  <j>  =  tan-1  m,  and  since  ds2  =  dx2  +  dy2; 

dd>  =  =— ; — ;    and  ds  =  Vl  +  m2  dx,  where  m  =  -¥- ; 

v      1  +  m2  dx ' 

hence  the  radius  of  curvature 

ds       Vl  +  m2  dx      (1+ra2)* 


R 


d(ji  dm  dm 

1  +  m2  dx 


(1  +m2)? 

b         ~  d?i 

dx 


(i) 


d2w 
#  will  be  positive  or  negative  according  as  -^  is  positive  or 

negative,  if  <f>  is  always  taken  as  the  acute  angle  which  the 
tangent  makes  with  the  x-axis;  for  then,  whether  $  is  posi- 
tive or  negative,    1  +  ( ~)       =  sec30  will  be  positive  and 

d2y 
the  sign  of  R  will  be  that  of  ■-— .     Hence  the  sign  of  R  will  be 

plus  or  minus  according  as  the  curve  at  the  point  is  concave 

sec3  cb 
upward  or  downward.     R  may  be  in  the  form  — r — • 

If  the  reciprocals  of  the  members  of  (1)  are  taken,  then 

K  -  t  =  3/t1 + (!)?'  which-may be  in  the  f0™ 

b  cos3  4>. 

d2v  1  . 

If  -~  is  zero  at  any  point  of  a  curve,  then  K  =  -^is  zero 

and  R  is  infinite.  Thus  at  a  point  of  inflexion  R  is  infinite. 
It  may  be  noted  that  as  a  curve  approaches  being  a  straight 
line,  its  curvature  approaches  zero  and  its  radius  of  curvature 
becomes  infinite,  that  is,  it  increases  in  length  without  limit. 
So  a  straight  line  is  the  line  that  the  arc  of  a  circle  of  curva- 


138  DIFFERENTIAL  CALCULUS 

ture  approaches  as  the  radius  of  the  circle  increases  without 
limit.  On  the  contrary  as  the  radius  of  curvature  at  a  point 
approaches  zero,  the  curvature  at  the  point  becomes  infinite 
and  the  curve  will  approach  a  mere  point,  since  the  circle  of 
curvature  will  diminish  with  zero  as  a  limit  for  its  radius. 
92.   Approximate  Formula  for  Radius  of  Curvature.  — 

Since  K  = 


['+©■] 


3' 


it  is  seen  that  the  flexion  when  multiplied  by  the  factor 
1/(1  +  m2)^  gives  a  measure  of  the  bending  of  a  curve 
independent  of  the  position  of  the  axes. 

The  flexion  is  the  rate  of  change  of  the  tangent  of  the  inclina- 
tion of  the  curve  at  a  point  with  respect  to  the  abscissa,  while 
the  curvature  is  the  rate  of  change  of  the  inclination  of  the 
curve  at  a  point  with  respect  to  the  arc,  where  the  inclination 
of  the  curve  is  that  of  the  tangent  line  at  the  point. 

However,  when  the  curve  deviates  but  slightly  from  a 
horizontal  straight  line,  the  curvature  is  approximately  the 

same  as  the  flexion,  since  the  slope  m  —  -j-  being  small,  f-p) 

is  very  small  compared  with  1,  and  therefore  the  formula 
becomes  approximately 

This  approximation  for  the  curvature  is  used  to  advantage 
in  the  flexure  of  beams  and  columns.  The  approximate 
formula  for  the  radius  of  curvature  is  consequently 

dx2 


EXERCISE  XII  139 

EXERCISE   XH. 

1.    To  find  R  and  the  curvature  of  the  ellipse  —  +  —  =1. 

a1      bl 

dy  _  _  fr^c        d?y  _        ¥ 
dx  a2y'      dx2  a2y3 

Substituting  these  values  in  (1),  Art.  91,  gives 

\    *aYJ   \       V  )  ^ ' 

v      d<t>      1  a464 


ds      R  (ay  +  5V)! 

The  maximum  curvature  is  a/b2,  at  A  (a,  0),  where  R  =  62/a  is  a 
minimum,  and  the  minimum  curvature  is  b/a2.  at  B  (0,  6),  where 
#  =  a2/b  is  a  maximum.     (See  Art.  90,  figure;  Art.  97,  figure.) 

2.  To  find  R  and  the  curvature  of  the  parabola  y2  =  4  px. 

dy  =  2_p        (Py  _  _  4p2 
ete        2/  '      dx2  y3 

Substituting  these  values  in  (1)  of  Art.  91  gives 

ds       B  2(x  +  p)§' 

At  the  vertex  (0,  0),  R  =  2  p,  the  minimum  radius;  and  the  maxi- 
mum curvature  is  (1/2  p)  radian  to  a  unit  of  arc.  (See  Example  1, 
Art.  97.) 

d2y          4  »2 
Since    -r^  = j-  is  negative  for  positive  values  of  y  and  posi- 
tive for  negative  values,  the  curve  is  concave  downward  at  points 
whose  ordinates  are  positive,  and  concave  upward  at  points  whose 
ordinates  are  negative.     The  sign  of  R  may  be  neglected,  since  the 

sign  of  -j~  will  indicate  whether  the  curve  is  concave  upward  or  down- 
ward at  any  point .  

3.  To  find  R  of  the  cycloid  x  =  a  vers-1  (y/a)  T  V2  ay  —  y2. 

dy  =  V2  ay  -  y2        d2y  =       a  t 
dx  y         '      dx2  y2 


140  DIFFERENTIAL  CALCULUS 

Substituting  these  values  in  (1)  of  Art.  75  gives 

-(i+sVJff),(-9-C^,(-S)-*VI* 

At  the  highest  point,  y  =  2  a,  and,  therefore,  R  =  4  a,  the  maximum  of 
R.     At  the  vertex  (0,  0),  R  =  0,  and  also  at  other  points  where  y  is 
zero;    therefore,  R  being  zero,  at  those  points,  which  are  cusps,  the 
curvature  is  infinite.     (See  Example  3,  Art.  97,  figure.) 
Find  R  and  the  curvature  of  each  of  the  following  curves. 

4.   The  equilateral  hyperbola  2  xy  =  a2.        R  =  (x2  +  y2)l/a2. 


5.  The  cubical  parabola  yz  =  a?x. 

6.  The  logarithmic  curve  y  =  bx. 

7.  The  catenary  y  =  |  (ex/a+  e~x/a). 

8.  The  hypocycloid  x*  +  y3  =  a3. 

9.  The  curve  x*  +  ?/*  =  ai 

93.   Radius  of  Curvature  in  Polar 

(4)  and  (10)  of  Art.  67, 

0  =  A  _j_  ^     ^  =  tan-1 


d<t>  6  a4*/ 


JV    - 

~  ds 

(9  yi+  a4)f 

d<j> 

my 

ds 

{in 

2  +  y2)* 

d4> 
ds 

a 

R  = 

■■  3  (axy)*. 

ds 

a- 

2  (.f  +  y)j 

Coor 

ding 

ites.  —  From 

p 

c/p/d0' 


d4  =         cbp       cU  =  (c/p/^)2  -  p  •  d2p/rffl2 . 
""'      dd  "     +  d0 '      c/0  p2  +  (dp/d0)2 

d^_  p2  +  2  (dp/dfl)2  -  p  •  d2p/dd\ 
'""      d0  ~  p2  +  (dp/d0)2 

.  ds/d0  _  [p2  +  (dp/dd)^ ,  , 

*  *    a    d<f>/dd    p^  +  2  (rfp/^)2  -  P  •  dr-p/de3 '  w 

Corollary.  —  Since  R  =  oo  at  a  point  of  inflexion, 
p?  +  2(dp/d0)*-p.g=O 
is  a  necessary  condition  for  such  a  point. 


RADIUS  OF  CURVATURE  IN  POLAR  COORDINATES  141 


Example.  —  To  find  R  for  the  curve  p  =  sin  0.     Here 

dp  a      d2P  •    a 


R 


(p2  +  COS20)* 


(sin2  0  +  cos2  0) 


1 


p2+2cos20-p(-sin0)      sin20+2cos20+sin20      2 

This  curve  p  =  sin  0,  a  circle  with 
unit  diameter,  in  connection  with  the 
formula    for    polar    curves,    tan  \f/  = 

—jr,  furnishes  a  derivation  of   the 


dp/dd 

d  (sin  6) .     Since  for  circle 


yp  =  d,    tan  6  = 


dp/dd 


dp 


sin0 


tan  6      tan  6 

that  is,        d  (sin  6)  =  cos  6  dO. 
Also  from  figure: 

OP         P 


dd  =  cosddd; 


tan0 


OA      cos0' 


and  since 


tan0  = 

dp 

dd  = 


dp/dd' 


cos  0  =  subnormal  OA ; 


so  again,     d  (sin  0)  =  cos  0  dd. 

This  curve  serves  as  an  illustration  of  maxima  and  minima 
in  polar  coordinates.     Thus,  p  =  sin  0  will  be  a  maximum 


or  a  minimum  when 


dp 


=  cos  0  =  0, 


when  0  =  -  or 


3tt 


and    since 


dd2 


TV 

—  sin  0,   is   negative   when    0  =  s 


<22p 


p  =  sin  ;r  =  1  is  a  maximum,  while 
r  2  «02 


—  sin  0  is  posi- 


142 


DIFFERENTIAL  CALCULUS 


trve  when  0  = 


3tt 
2  ' 


p  =  sin  —  =  —  1  is  a  minimum. 


As  the  denominator  of  the  fractional  value  of  R  is  2  for  any 
value  of  0,  there  is  no  inflexion  point,  R  not  being  infinite 
at  any  point. 

EXERCISE   Xm. 

Find  R  in  each  of  the  following  curves : 


1.  The  Cardioid  p  =  a  (1  —  cos0). 

2.  The  Lemniscate  p2  =  a2  cos  2  0. 
Where  is  an  inflexion  point  ? 

3.  The  Spiral  of  Archimedes  p  =  a 

4.  The  Logarithmic  Spiral  p  =  ae. 


R  =  2V2aP/3. 
fl  =  a2/3  p. 


R  =  a 


(d2  +  D§ 
02  +  2 


R  =pVl  +  (log  a)2. 


94.   Coordinates  of  Center  of  Curvature.  —  Let  P  (x,  y) 
be  any  point  on  the  curve  ab,  and  C  (a,  0)  the  corresponding 


center  of  curvature.     Then  PC  is  R  and  is  perpendicular  to 
the  tangent  PT.     Hence 

Z  BCP  =  z  XTP  =  4>, 

OA  =  OM  -  BP,    AC  =  MP  +  BC; 

that  is,  a  =  x  —  R  sin  cf>,     (3  =  y  -f  R  cos  0; 


or 


a  =  x 


Rfs-     f 


.     r>dx 


(i) 

(2) 


PROPERTIES  OF  THE  INVOLUTE  AND  EVOLUTE   143 


Substituting  in  (2) .  the  values  of  R  and  ds,  gives 


a  =  x  — 


dx2 


0  =  2/  + 


dx2 


(3) 


95.  Evolutes  and  Involutes.  —  Every  point  of  a  curve,  as 
in,  has  a  corresponding  center  of  curvature.  As  the  point 
(x,  y)  moves  along  the  curve  in,  by  equation  (3)  above,  the 


point  (a,  /3)  wall  trace  another  curve,  as  ev.  The  curve  ev, 
which  is  the  locus  of  the  centers  of  curvature  of  in,  is  called 
the  evolute  of  in. 

To  express  the  inverse  relation,  in  is  called  an  involute  of 
ev.     The  figure  shows  an  arc  of  an  involute  of  a  circle. 

96.   Properties  of  the  Involute  and  Evolute.  —  I.  Since 

dx/ds  =  cos  4>,     dy/ds  =  sin  </>,     and     ds  =  R  d(j>, 

dx  =  cos  4>ds  =  R  cos  <f>  d<f>,  (1) 

and  dy  =  sin  0  ds  =  R  sin  0  d<p.  (2) 

Differentiating  equations  (1)  of  Art.  94,  and  using  the 
relations  given  in  (1)  and  (2),  there  results 

da  =  dx  —  R  cos  0  d(f>  —  sin  4>  dR  =  —  sin  <j>  dR,         (3) 
d(3  =  dy  -  R  sin  0  d<t>  +  cos  </>  dR  =  cos  <f>  dR.  (4) 


144  DIFFERENTIAL  CALCULUS 

Dividing  (4)  by  (3)  gives 

d(3/da  —  —  cot  0  =  —  dx/dy. 

That  is,  the  normal  to  the  involute  at  (x,  y),asP  (Art.  95,  figure), 
is  tangent  to  the  evolute  at  the  corresponding  point  (a,  (3),  as  C. 
II.   Squaring  and  adding  (3)  and  (4)  gives 
da2  +  dp2  =  dR2. 
Let  s  denote  the  length  of  an  arc  of  the  evolute;  then, 

da2  +  dp2  =  ds2. 
Hence,  ds  =  ±dR;     :.     As  =  ±  A#. 

That  is,  the  difference  between  two  radii  of  curvature,  as 
C3P3  and  C1P1  (Art.  95,  figure),  is  equal  to  the  corresponding 
arc  of  the  evolute,  C1C3. 

These  two  properties  show  that  the  involute  in  can  be 
described  by  a  point  in  a  string  unwound  from  the  evolute 
ev.     From  this  property  the  evolute  receives  its  name. 

It  may  be  noted  that  a  curve  has  only  one  evolute,  but 
an  unlimited  number  of  involutes,  as  each  point  on  the  string 
which  is  unwound  would  describe  an  involute. 

97.  To  Find  the  Equation  of  the  Evolute  of  a  Given 
Curve.  —  Differentiating  the  equation  of  the  given  curve  and 
using  equations  (3)  of  Art.  94,  a  and  0  will  be  expressed  in 
terms  of  x  and  y.  These  two  equations  and  that  of  the  given 
curve  furnish  three  equations  between  a,  jS,  x  and  y.  Elimi- 
nating x  and  y  from  these  equations,  a  relation  between  a 
and  jS  is  obtained,  and  this  relation  is  the  equation  of  the 
evolute  of  the  given  curve,  which  would  itself  be  an  involute  of 
the  curve  found. 

Examples.  —  Find  the  equation  of  the  evolute  of  the 
following  curves: 

1.   The  parabola  y2  =  4px.  (1) 

dy  _  2  p       d?y  _       4p2 
dx       y  '     dx2  y3 


EQUATION  OF  THE  EVOLUTE 

Substituting  these  values  in  (3)  of  Art.  94  gives 
a  =  Sx  +  2p;    0=  -y*/±p2; 
/.     x  -  (a  -  2  p)/3,     7/  =  -  ^402? 
Substituting  these  values  of  z  and  y  in  (1)  gives 

/32  =  4(a-2p)3/27p, 

as  the  equation  of  the  evolute  of  y2  =  4  px. 

The  locus  of  (2)  is  the  semi- 
cubical  parabola.  Thus,  if 
iOn  is  the  locus  of  (1),  F 
being  the  focus,  then  eAv  is 
the  locus  of  (2) ,  where  OA  = 
2  p  =  2  OF  is  the  minimum 
radius  of  curvature  at  0,  the 
point  of  maximum  curvature 
on  the  parabola.  (Example 
2,  Exercise  XII.) 

2.   The  ellipse 
a2y2  +  &V  =  a262.     (1) 

aj/  _   _  b2x     d2y  _        64 
dx  a2y}    dx2  a2yz 

Substituting  these  values  in 
(3)  of  Art.  94  gives 

(a2  -  }f)  x* 


145 


(2) 


a  = 


0  = 


(a2-62)i/3 


\a2  -  b2)  '     y        \a2  -  b2) 


Hence,  the  equation  of  the  evolute  of  the  ellipse  a2y2  +  b2x2  = 
a2b2  is 

(aa)3  +  (6/3)i  =  (a2  -  62)l 

The  evolute  is  C1C2C1C2.    Ci  is  center  of  curvature  for 


146 


DIFFERENTIAL  CALCULUS 


A;  C  for  P;  C2  for  B;  &'  for  A';  C2'  for  B'.  In  the  figure 
shown  a  =  2b;  when  a  =  6  V2,  then  the  center  of  curvature 
for  B  is  at  B'  and  for  B'  at  £.     When  a  <  b  V2,  the  centers 

for  B  and  B'  are  within 
the  ellipse.  The  points  Ci, 
C2,  CV,  and  C2  are  cusps. 
The  length  of  the  evolute 
is  evidently  four  times  the 
difference  between  R  at 
B  (a,  b),  and  R  at  A  (a,  0); 
that  is,  (1,  Exercise  XII), 
4(a2/6-&2/a)  =4(a3-63) 
/a&. 

Corollary.  —  For  circle, 
since  a  =  b,  the  evolute  is 
a  point,  the  center  of  the 
circle. 


The  involute  of  the  circle  is 

given  by  the  equations,  0     n  a 

x  =  a  (cos0  +  0sin0), 

ij  =  a  (sin0  —  0cos0). 

AP  is  the  arc  of  an  involute  of  the  circle. 

3.   The  cycloid  x  —  a  vers-1  (y/a)  =F  ^2  ay  —  y2. 

dy  _  V2  ay 
dx 


M 


//- 


dx2 


y  ax"  y 

Substituting  these  values  in  (3)  of  Art.  94: 
y  =  -  ft     x  =  a  =  2  V  -  2  a/3 
.*.     a  =  a  vers-1  ( —  j3/a)  d=  V 


/32; 


2ap-0*.  (1) 

The  locus  of  (1)  is  another  cycloid  equal  to  the  given  one,  the 


EQUATION  OF  THE  EVOLUTE 


147 


highest  point  being  at  the  origin;    that  is,  the  evolute  of  a 
cycloid  is  an  equal  cycloid.     Thus,  the  evolute  of  the  arc  OP\ 
is  the  arc  OCi,  which  equals  Pix;  and  the  evolute  of  P\X  is 
Cix,  which  equals  OP\. 
Since  R  =  2  V2ay  (3,  Exercise  XII),  CiPi  =  4 a.    Then 
OPxX  =  2  •  OCi  =  2  •  CiPi  =  8a. 

Hence,  the  length  of  one  branch  of  the  cycloid  is  eight  times 
the  radius  of  the  generating  circle.  (See  Example  3,  Art. 
142.) 


If  the  figure  shown  be  inverted,  the  principle  of  the  cy- 
cloidal  pendulum  may  be  perceived.  A  weight,  suspended 
from  G  by  a  flexible  cord,  may  be  made  to  oscillate  in  the 
arc  OPiX,  by  means  of  some  surface  shaped  like  the  arcs 
CiO  and  C\X  causing  a  continuous  change  in  the  length  of  the 
cord  as  it  comes  in  contact  with  the  surface.  The  cycloidal 
pendulum  is  isochronal,  as  the  time  of  an  oscillation  is 
independent  of  the  length  of  the  arc.     (See  Art.  237.) 

4.  The  hyperbola  b2x2  —  a2y2  =  a2b2. 

(aa)3  -  (&/3)*  =  (a2  +  b2)*. 

5.  The  equilateral  hyperbola  2  xy  =  a2. 

(a  +  (3p  -  (a  -  /3)3  =  2  a*. 

6.  Find  the  length  of  an  arc  of  the  evolute  of  the  parabola 
y2  —  4  px  in  terms  of  the  abscissas  of  its  extremities. 


148  DIFFERENTIAL  CALCULUS 

Arc  AC  =CP-AO  =  2(*  +  P)s  _  2p    (Example  1,  figure) 

7.  Show  that  in  the  catenary  y  =  a/2  \ea  +  e  a), 

a  =  x  —  y/a  Vy2  —  a2,     (3  =  2y. 

8.  Find  the  equation  of  the  evolute  of  the  hypocycloid 

XZ  -J-  y3  =  az, 

(a  +  0)*  +  (a-/3)*  =  2  a*. 


CHAPTER  VII. 

CHANGE   OF   THE   INDEPENDENT   VARIABLE. 
FUNCTIONS  OF  TWO  OR  MORE  VARIABLES. 

98.   Different  Forms   of   Successive   Derivatives.  —  As 

given  in  Arts.  67,  68,  where  x  is  independent  dx  may  be 
taken  as  having  always  the  same  value  and  is  accordingly 
treated  as  a  constant;  hence, 

d  dy  _  d2y       d_   d   dy  _  d3y 

dx  dx      dx2'    dx  dx  dx      dx3' 

When  neither  x  nor  y  is  independent,  -j-  is  a  fraction  with 

both  numerator  and  denominator  variable,  and  d  dx  =  d2x} 
etc.,  hence, 

d  dy  _  dx  d2y  —  dy  d2x  ,  . 

dx"dx~  dx*  '  {) 

d_d_dy_  dx2  dzy  -  dx  dy  dzx  -3dx  d2x  d2y  +  3dy  (d2x)2 
dx  dx  dx  dxb  ' 


When  y  is  independent,  d2y  =  0,  d3y  =  0,  .  .  .   ;   hence, 
d  dy  _       dy  dzx 
dx  dx  dx3    ' 

d    d  dy      3  dy  (d2x)2  —  dx  dy  d?x 


dx  dx  dx  dx1 


do 

(2') 


99.  Change  of  the  Independent  Variable.  —  In  some  ap- 
plications of  the  Calculus  it  is  necessary  to  make  a  differen- 
tial equation  depend  on  a  new  independent  variable  in  place 
of  the  one  originally  selected;  that  is,  there  is  need  to  change 
the  independent  variable. 

149 


150  DIFFERENTIAL  CALCULUS 

When  x  =  4>(z)  and  it  is  desired  to  change  the  independent 

d~y    d3y 
variable  from  x  to  z;  for  -r\,  -r\,  .  .  .  ,  respectively,  the 

second  members  of  (1),  (2),  .  .  .  ,  above,  are  substituted; 

and  in  the  resulting  equation,  for  x,  dx,  d2x,  .  .  .  ,  their 

values  gotten  from  the  equation  x  =  4>  (z)  are  substituted. 

Example  1.  —  Given  y  d2y  +  dy2  +  dx2  =  0,  in  which  x  is 

independent,   to  find   the  transformed  equation  in  which 

neither  x  nor  y  is  independent;    also  the  one  in  which  y  is 

independent. 

d2y 
Dividing  both  members  by  dx2,  substituting  for  -r^  the 

second  member  of  (1)  Art.  98,  and  multiplying  both  mem- 
bers by  dx3,  gives 

y  (d2y  dx  —  d2x  dy)  +  (dy2  dx  +  dx3)  =  0, 

in  which  neither  x  nor  y  is  independent. 

Putting  d2y  =  0,  and  dividing  by  —  dy3,  gives 

d2x      dx3      dx  __  n 
yay~ay~dy~    ' 

in  which  the  position  of  dy  indicates  that  y  is  independent. 
Example  2.  —  To  change  the  independent  variable  from 
x  to  6  in 

p      [l  +  (dy/dx)2}\ 
d?y 
dx2 

given  x  =  p cos 6,y  =  p sin 0,  p  being  a  function  of  6. 
From  the  data, 

,  dy  =  sin  0  dp  +  p  cos  0  tf#, 
dx  =  cos  0  dp  —  p  sin  0  dd, 
d2y  =  sin  6  d2p  +  2  cos  ddddp-  p  sin  0  d02, 
and 

d2x  =  cos  dd2p-  2  sin  ddddp-  p  cos  (9  d02. 


EXERCISE  XIV  151 

Substituting  these  values  in  value  of  R  and  simplifying,  gives 

=    \f  +  (dp/dey]t     the  ^^  o{  ^  .n  ^  9^ 

To  change  the  independent  variable  from  x  to  i/;  for 
d2y/dx2,  (Py/dx*,  .  .  .  ,  respectively,  the  second  members  of 
(1'),  (2'),  •  •  •  ,  above,  are  substituted;  or  in  the  general 
result,  as  in  example  1,  make  d2y  =  0,  d?y  =  0,  etc. 

Example  3.  —  Change  the  independent  variable  from  x  to 
y  in 

3AW  _dy<$y_  d2y/dy\2  =  Q 
\dx2)       dx  dx?      dx2  \dx) 

d2v  d^v 

Substituting  for  -r\  and  -r\ ,  respectively,  the  second  mem- 
bers of  (1')  and  (2')  gives  after  reduction 
d3x      d2x  _  n 

¥3  +  ¥2"  ' 

in  which  the  position  of  dy  shows  that  y  is  independent. 


EXERCISE   XIV. 

1.  Given  x  =  cos  9,  change  the  independent  variable  from  x  to  0  in 

i& ^_  ^y  ,      y     _  0  A       d2y        _ 

dx2      1  -  x*  dx  +  1  -  x>  -  U*  Ari^    d9»  +  y  ~  °' 

2.  Given  x  =  -,  change  the  independent  variable  from  z  to  z  in 

3.2^4.0x^4-^7;  -0  y1r?9     ^4_„2„_o 

*   e^  +  ^dx  +  x2*'-0-  An5,    ^+a^/-°- 

3.  Given  x2  =  4  2,  change  the  independent  variable  from  x  to  2  in 

4.  Given  x  =  cos  2,  change  the  independent  variable  from  x  to  2  in 


152  DIFFERENTIAL  CALCULUS 

5.  Change  the  independent  variable  from  x  to  y  in 

g +  «,--. )(g)*  =  o.  a*  J*.-* 

6.  Given  2  =      ,     ,       ,   ,  to  find  the  transformed  equation  when 

ydy  -\-xdx' 

x  =  p  cos  6,  y  =  p  sin  0,  and  p  is  independent.  Arcs,  z  =  p  —  - 

dp 

100.  Function  of  Several  Variables.  —  A  function  may 
depend  upon  two  or  more  variables  having  no  mutual  rela- 
tion, that  is,  independent  of  each  other.  Thus  the  volume 
of  a  gas  depends  upon  the  temperature  and  also  upon  the 
pressure  to  which  it  is  subjected,  and  the  temperature  and 
the  pressure  may  vary  independently. 

A  variable  z  is  a  function  of  the  independent  variables  x, 
y,  .  .  .  when  for  each  set  of  values  of  these  variables  there 
is  determined  a  definite  value  or  values  of  z. 

A  function  of  two  variables 

z  —  f  (x,  y),  where  x  and  y  are  independent, 

is  represented  geometrically  by  a  surface,  plane  or  curved 
according  to  the  form  of  the  functional  relation ;  and  to  each 
pair  of  values  of  (x,  y)  there  corresponds  a  point  on  this 
surface.  When  x  and  y  vary,  the  point  takes  another  posi- 
tion, and  it  will  take  the  new  position  either  by  x  and  y 
varying  simultaneously  or  by  one  remaining  constant  while 
the  other  changes. 

101.  Partial  Differentials.  —  A  partial  differential  of  a 
function  of  two  or  more  variables  is  the  differential  when 
only  one  of  the  variables  is  supposed  to  change.  Let  z  = 
f  (x,  y)  be  the  surface  shown  in  the  figure,  and  P(x,  y,  z)  the 
moving  point;  then  if  y  is  constant  while  x  changes,  P  will 
move  on  the  plane  curve  PA  and  dx  may  be  represented  by 
PM  or  P'M'-y  on  the  other  hand,  if  x  is  constant  while  y 
changes,  P  will  move  on  the  plane  curve  PB  and  dy  may  be 
represented  by   PN  or    P'N'.    The  differential  of  2  as  a 


PARTIAL  DIFFERENTIALS 


153 


function  of  x,  y  being  regarded  as  a  constant,  is  denoted  by 
dxz;  and  the  differential  of  z  when  y  alone  is  variable  is 
denoted  by  dyz.  These  differentials  are  the  partial  differ- 
entials of  z  with  respect  to  x  and  y,  respectively. 


Note.  —  The  partial  dxz  may  in  the  figure  be  represented 
by  the  distance  on  the  ordinate  from  the  point  M  to  the 
tangent  TP,  and  so  too,  the  partial  dyz  may  be  represented 
by  the  distance  from  N  to  the  tangent  T'P,  both  being 
negative  in  this  case  as  z  is  decreasing.  The  A^  and  the 
Avz  are  the  distances  from  the  points  M  and  N  to  the  surface 
curves  through  P2. 

102.   Partial  Derivatives.  —  The  partial  derivatives  of  z 

with  respect  to  x  and  y  are  denoted  by  ■=-  and  -7- ,  respectively, 

and  they  may  be  represented  by  the  equivalent  notation, 
fx'(x,  y)  and  fv'(x,  y). 

In  the  figure  of  Art.  101,  consider  P  as  the  intersection  of 
the  curves  CPA  and  C'PB,  cut  from  the  surface  by  the 
planes  y  =  b  and  x  =  a,  respectively;  then  the  slope  of  the 


154  DIFFERENTIAL  CALCULUS 

dz 

curve  CPA  is  given  by  the  partial  derivative  -7- ,  and  that  of 

the  curve  C'PB  by  the  partial  derivative  -p ;  that  is,  the 

partial  derivatives  are  the  tangents  of  the  inclination  of  the 
tangent  lines  at  P  to  the  axes  of  X  and  Y,  respectively.  The 
values  of  the  slopes  for  some  definite  point  P  on  the  surface 

are  gotten  by  substituting  in  the  expressions  for  -r-  and  -j- , 

respectively,  the  corresponding  values  of  x  and  y.  Thus  in 
this  case  (a,  6),  or  P' ,  being  the  projection  of  P  on  the  xy- 
plane,  a  is  substituted  for  x  and  b  for  y. 

103.  Tangent  Plane.  Angles  with  Coordinate  Planes. 
—  In  the  figure  of  Art.  101,  let  P  be  the  point  (xh  yh  Zi) ; 
PT,  the  tangent  to  CPA  in  the  plane  y  =  y\\  and  PT ',  the 
tangent  to  C'PB  in  the  plane  x  =  Xi. 

The  equations  of  PT  are 


Sl  =  [fi(a 


si)i  y  =  Vh  (i) 

and  of  PT', 

z-zl  =  L^J  (?/  -  2/1),   z  =  zi.  (2) 

The  plane  tangent  to  the  surface  at  P  has  for  its  equation, 

-*-[£l«»J*a +[*!<*-»>•       (3) 

since  it  is  determined  by  the  two  intersecting  tangents,  is 
of  the  first  degree  with  respect  to  x,  y,  z,  and  is  satisfied  by 
(1)  and  (2). 

The  equations  of  the  normal  through  P  are  those  of  a  line 
through  (xh  yh  zx)  perpendicular  to  (3).     Its  equations  are 

-* /[bI-'VKI--*-*-   (4) 

The  angles  made  by  the  tangent  plane  with  the  coordinate  planes 
are  equal  to  the  inclinations  of  the  normal  to  the  axes. 


TANGENT  PLANE  155 

The  direction  cosines  of  the  line  perpendicular  to  (3)  are 

proportional  to  [|],  [g],-l. 

Hence,  if  a,  (3,  y,  are  the  inclinations  of  the  normal  to  OX, 
OY,  OZ,  respectively, 

°"B/[sl =  ""/[SI =  C0ST/"  l      (5) 

Also  cos5  a  +  cos2  /3  +  cos2  7  =  1.  (6) 

From  (5)  and  (6),  in  general,  at  any  point  (x,  y,  z), 

— ©r/'+e?+eif- 

-MS)V'+(£HI)" 

For  the  inclination  of  the  tangent  plane  to  XY,  from  (7), 

— ©T+6ST-  (9) 

From   (9),  calling  the  tangent  of  the  angle  made  by  the 
tangent  plane  with  the  plane  XY  the  slope, 


**--  vewt)* 

Example.  —  Find  the  equations  of  the  tangent  plane  and 
normal,  to  the  sphere  x2  +  y2  +  z2  =  a2,  at  (xh  yh  2i). 

3«  _  __ x       $i  -  _y. 

dx~       z'  dy~       z' 

|"^1  -  _  £!      r^il  =  -y-l. 


156  DIFFERENTIAL  CALCULUS 

Substituting  in  (3), 

z-Zi  =  -  f*  (x  -  xi)  -  v-r  (y  -  2/1), 
xxi  +  yyi  +  zzi  =  xx2  +  yx2  +  zY2  =  a2.  Arts. 
From  (4)  for  the  normal : 

(x  -  xi)  ^  =  (y  -  yi)  ^  =  z  -  si, 

xi  yi  Z\  xi      2/1      zi 

EXERCISE   XV. 

1.  Find  the  equation  of  the  tangent  plane  and  its  slope,  for  the 
ellipsoid,  x2  +  2j/8  +  3z2  =  20,  at  (3,  2,  +z,). 

Arcs.  3  x  +  4  2/  +  3  z  =  20;  f. 

2.  Find  the  equation  of  the  tangent  plane  to  the  elliptic  paraboloid, 
z  =  3  x2  +  2 1/2,  at  the  point  (1,  2,  11). 

Ans.  6  x  -\-  8y  —  z  =  11. 

3.  Find  the  equations  of  the  tangent  plane  and  normal  to  the  cone, 

3x2  -2/2  +  2z2  =0,  at  {xhyhzx). 

a        o  10  »    x  -  xx      y  -  yx      z  -  Zi 

Ans.  3  xxi  -2/2/1  +  2  ggi  =  0;  =  - — -f-  =  -x- — 

0X1  —2/i  Z  Zi 

iVote.  —  The  equations  of  the  tangent  plane  and  normal 
are  illusory  if  formed  for  the  origin.  Every  tangent  plane 
to  the  cone  goes  through  the  origin  and  there  is  no  definite 
normal  at  the  origin.  When  at  special  points  on  a  surface 
the  three  partial  derivatives  of  the  function  with  respect  to 
each  of  the  three  variables  are  all  zero,  there  is  no  definite 
tangent  plane  or  normal  at  the  point.  Such  points  are 
called  conical  points,  the  vertex  of  a  cone  being  the  typical 
case. 

104.  Total  Differentials.  —  When  z  =  f  (x,  y)  is  differ- 
entiated, both  x  and  y  varying,  the  total  differential  dz  or 
df  (x,  y)  is  gotten. 

The  derivations  of  the  formulas  for  differentiation  of  al- 
gebraic,  logarithmic   and   exponential   functions,   given  *in 


TOTAL  DIFFERENTIALS  157 

Chapter  II,  hold  when  u,  v,  y,  and  z  denote  functions  of  two 
or  more  independent  variables;  hence  the  total  differential 
of  /  (x,  y)  may  be  gotten  by  the  principles  established  in 
those  derivations.  The  total  differential  of  a  function  of  two 
or  more  variables  is  equal  to  the  sum  of  its  partial  differentials. 
If  z  =  /  (x,  y) ,  then 

dz  =  dxz  +  dyz  =  -^dx  +  ^.dy; 
and  if  v  =  fix,  y,  z),  then, 

dv  =  dj)  +  dvv  +  dzv  =  -j-  dx  +  -7-  dy  -f-  -5-  dz> 
y  dx  dy   u      dz 

where  the  last  form  of  the  partial  differentials  is  another 
convenient  notation.  In  the  figure  of  Art.  101,  dz  is  repre- 
sented by  the  distance  on  the  ordinate  from  D  to  the  tan- 
gent plane  at  P  and  is  there  negative,  Iz  being  DP2,  which 
is  negative. 

The  truth  of  this  theorem  has  been  illustrated  geometrically 
in  the  derivations  of  d(uy)  and  d(xyz)  in  Arts.  28  and  29,  and 
the  theorem  is  readily  established  analytically.  Thus,  it  has 
been  found  that  all  the  terms  of  d(f(x,  y))  are  of  the  first 
degree  in  dx  and  dy;  hence,  if  z  =  f  (x,  y), 

dz  =  <f>(x,  y)  dx  +  4>i(x,  y)  dy,  (1) 

where  <p(x,  y)  and  4>\{x,  y)  denote,  respectively,  the  sums  of 
the  coefficients  of  dx  and  dy  in  the  several  terms  of  dz. 
When  x  alone  varies,  (1)  becomes 

dxz  =  <f>  (x,  y)  dx.  (2) 

When  y  alone  varies,  (1)  becomes 

dyz  =  0i  (x,  y)  dy.  (3) 

Hence,  from  (1),  (2)  and  (3), 

dz  =  dxz  +  dyz  r=  -£dx  +^dy. 


158  DIFFERENTIAL  CALCULUS 

105.  If  .  =  ,(,,,)  =  c,    *  -g*  (!) 

for  j|<te  +  g<%  =  <fe  =  d(c)=0,  (2) 

which  solved  for  -p  gives  (1). 

This  formula  for  the  derivative  of  an  implicit  function  is 
useful  in  many  cases. 
Example.  —  Given  x4  —  a2xy  +  Vy2  =  c  =  z,  to  find  dy/dx. 

Here         -=-  =  4  £3  —  a2y     and     -=-  =  —  a2x  +  2  62?/; 

dy       4  a;3  —  a2w  ,       ,    '     N 


EXERCISE   XVI. 

1.  u  =  x2/a2  +  2/V&2- 

dw  _  2x         3m  _  2y 
dx~  a2  '        dy~  ~W' 
£du.y  du  _  x2      y2  _ 
"      2  dx  +  2  dy~  ~  a2  +  P  ~  U' 

2.  w  =  b  X  y'  +  ex2  +  eyz. 

d£  =  by2  +  2cx,     ^  =  2bxy+Sey2. 

n  xy  du   .         du 

3.  u  = — = —  x*-r-+y-j- =  u. 

x  +  y  dx  dy 

AX  du 

4-u=2/-  Tx 


+  (|).|  =  «0ogv  +  D. 


6.  w  =  6xz/2  +  ex2  +  ei/3.        dw  =  (by2  +  2  ex)  dx  +  (2  &n/  +  3  ey2)  dy. 

7.  u  =  yx.  du  =  yx  logydx  +  xy1'1  dy. 

8.  U  =  l/sin  *.  du  —  ?/sin  *  log  y  COS  X  dx  +  7/sini-i  sin  x  dy. 

By  Art.  105,  find  dy/dx  when: 

9.  *+»>-3a*j,-0.  ^rfj^ag. 

ax      ax  —  y- 

10.  a*  -  y*  =  0.  ^  =  ?/~^logy- 

dx      x2  —  x?/  log  x 


TOTAL   DERIVATIVES  159 

n.  *iog»-»iog*-o.  *v.  =  y.***v-v. 

°  *       v     °  dx      x    y  log  x  —  x 

106.  Total  Derivatives.  —  If  u  =  f  (x,  y,  z),  y  =  <f>  (x), 
and  z  =  0i  (x),  u  is  directly  a  function  of  x  and  indirectly  a 
function  of  x  through  y  and  2.     The  total  differential, 

du  =  -r-dx  +-j-dy  +  -r-  dz     (by  Art.  104) 

becomes  by  dividing  by  dx, 

du  _  du      du  dy      du  dz  n . 

dx       dx      dy  dx       dz  dx' 

du 
where  -=-  is  the  total  derivative  of  u  as  a  function  of  x. 
dx 

Corollary  1.  —  If  u  =  f  (y,  z),  y  =  <l>(x),  and  z  =  #1(2), 

du  _  du  dy      du  dz  ... 

dx      dz/  dx      d?  dx 

Corollary  2.  —  If  u  =  f  (y)  and  y  =  0(x), 

du  _  du  dy  ,  . 

dx      dy  dx' 

where  -7-  is  the  derivative  of  a  function  of  a  function,  and 
dx 

(3)  is  the  formula  that  is  the  subject  of  Remarks  in  Art.  19. 

Corollary  3.  —  If  u  =  f  (x,  y,  z)  and  x,  y  and  z  are  inde- 
pendent of  each  other,  they  may  be  regarded  as  functions 
of  time  t;  hence,  the  expression  for  the  total  differential  du 
above  becomes  by  dividing  by  dt, 

du  _  du  dx      du  dy      du  dz  ... 

du 
where  -7-  is  the  total  time-derivative  or  rate  of  change  of  u. 

Similarly,  when  z  =  f  (x,  y),  x  and  y  being  functions  of  t, 

dz  _dz_  dx   .dz_  dy  ,_. 

dt~dx  ~dt+dy~dt'  ^  ' 


160  DIFFERENTIAL  CALCULUS 

If  y  is  a  function  of  x,  as  y  =  4>{x),  putting  x  for  t  in  (5), 
gives 

dx      ete      cfa/  eta 

107.  Illustrative  Examples.  —  Example  1.  —  The  edges 
of  a  right  parallelopiped  are  6,  8  and  10  feet.  They  are 
increasing  at  the  rate  of  0.02  foot  per  second,  0.03  foot  per 
second  and  0.04  foot  per  second,  respectively.  Show  at 
what  rate  the  volume  is  increasing. 

Let  volume  —  u  =  xyz,  then  by  (4),  Art.  106: 
du  dx  ,       dy  .        dz 

-17  =  VZ  -77  +  XZ  -77  +  XV  -T7 

dt       u   dt  dt        *  dt 

=  80  X  0.02  +  60  X  0.03  +  48  X  0.04 

=  1.60  +  1.80  +  1.92  =  5.32  cubic  feet  per  second. 

See  Art.  29  where  du  (=  dV)  is  shown  geometrically  by 
figure. 

Example  2.  —  Given  the  formula  for  gas,  pV  =  KT, 
where  p  is  pressure,  V  is  volume,  T  is  temperature,  and  K  is 
a  constant.  Let  K  =  50,  and  let  the  volume  and  tempera- 
ture at  a  given  time  be  V0  =  5  cu.  ft.  and  T0  =  250°.  The 
corresponding  pressure  is 

po  =  5°  X-  25°  =  2500  lb.  per  sq.  ft. 
o 

If  in  this  state  the  temperature  is  rising  at  the  rate  of  0.5 
degree  per  minute  and  the  volume  is  increasing  at  the  rate 
of  0.2  cu.  ft.  per  minute,  required  the  rate  at  which  the 

T 

pressure  is  changing.     Here  p  =  50  ^> 

,  dp      50      dp  _ft  T 

whenCe  _£  =  y ,    -£__»_. 

Hence,  in  the  given  state, 


1       50 

T=T0 


*'l 


ILLUSTRATIVE  EXAMPLES  161 


n- 

V=V0 


and  %\=    -™^°  = 


Given  ^  =  0.5     and     ^  =  0.2. 

dt  at 

Then 

by  (5)  Art.  106;  that  is,  the  pressure  is  decreasing  at  the 

rate  of  95  lb.  per  sq.  ft.  per  min. 

x^        iP"        z^ 
Example  3.  —  A  point  on  an  ellipsoid  —  +  —  +  t^  =  1, 

in  the  position  x  =  3,  y  =  —  4,  moves  so  that  x  increases 
at  the  rate  of  t two  units  per  second,  while  y  decreases  at  the 
rate  of  three  units  per  second.  Find  the  rate  of  change  of  z. 
Here 


dz_ 7_x dz_ 7y 


y2 


dx  dy  _ 

~dt~A  ~dt~6> 

dz 14x 21  y 

dt  ~ 


36 


\/-S-S  »^-S-SJ 


(by  (5) 

Art.  106) 


3,  -4 


-r-  = 7=  units  per  sec,  the  rate  of  change  of  z. 

dt      '  15  vn 

EXERCISE   XVH. 

1.   w  =  z2  +  y3  +  zy,     2  =  sin  a:,     y  =  ex;     find  •=— ■ 

Ans.    ■r-  =  3e3I+«I  (sin  x  +  cos  x)  +  sin  2  x. 


2.   u  =  vx2  +  t/2,    ?/  =  rax  +  c. 


dw  _  (1  +  m2)  x  +  rac 
dx       Vx2  +  (rax  +  c)2 


162  DIFFERENTIAL  CALCULUS 


3.  u  =  sin-1  (y  —  z),    y  =  3  x,    z  =  4  x3. 

4.  u  =  tan"1  ^ ,    x2  +  y2  =  r2. 

du 
dx 

du 
dx 

du 
dx 

5.  it  =  log  (z  +  y),    y  =  Vx2  +  a2. 

Vi  -x2 
1 


Vr2 
1 


Vx2  +  a2 

6.  With  the  same  data  as  in  illustrative  Ex.  2,  when  the  pressure  of 
the  gas  is  increasing  at  the  rate  of  40  lb.  per  sq.  ft.  per  sec.  and  the 
temperature  is  falling  at  the  rate  of  1  degree  per  sec,  find  the  rate  of 
change  of  the  volume. 

A  dV  ai  ft 

Ans.  -rr  =  —  0.1  cu.  ft.  per  sec. 

7.  A  point  on  a  elliptic  paraboloid  z  =  2  x2  +  5  t/2,  in  the  position 
x  =  —3,y  =  l,  moves  so  that  the  rate  of  change  of  x  is  3  units  per  sec, 
and  that  of  y  is  2  units  per  sec     Find  the  rate  of  change  of  z. 

Ans.  -r  =  —16  units  per  sec 

108.   Approximate   Relative   Rates   and   Errors.  —  The 

method  of  Art.  41  for  rinding  the  errors  or  small  differences 
in  a  function,  due  to  slight  variations  or  inaccuracies  in  the 
independent  variable,  is  applicable  to  a  function  of  two  or 
more  variables.  Since  when  an  area  A  =  /  (x,  y),  the  rela- 
tive rate  of  increase  of  A  is 

dA      dA 

dl_,dy      fx  0,  y)      fy'(x,y)  =fx,y'(A) . 

A  "*"  A  ~        A     '"*"      A  A      ' 

hence,  AA^Az  +  ^Ay  (1) 

AA      dA  Ax  ,  dA  Ay  /n, 

and  "d^X  +  ^T  (2) 

are  approximate  relations.  When,  for  example,  the  area  of 
a  rectangle  is  given  by  A  =  xy,  and  therefore,  dA  =  xdy  + 
y  dx,  when  x  and  y  are  the  measurements  and  dx  and  dy  the 
errors  or  inaccuracies,  then  dA  gives  the  approximate  error 
in  area  due  to  the  errors  dx  and  dy. 


APPROXIMATE  RELATIVE  RATES  AND  ERRORS      163 

If  a  rectangle  is  laid  out  1000  ft.  on  one  side  and  100  ft. 
on  the  other,  and  the  tape  is  0.01  ft.  too  long;  then. by  (1) 

AA  =y>kx  +  x-ky  =  100  X  0.1  +  1000  X  0.01 
=  10.00  +  10.00  =  20  sq.ft. 

is  the  approximate  error  and  the  exact  error  is  20.001  sq.  ft., 
found  by  more  laborious  computation.  The  approximate 
relative  error  is  by  (2) 

AA  =       20  1 

A   "  100,000"  5000' 

making  the  percentage  error 

100  AA        1  ft  no    *  i 

-. —  =  ptt    or    0.02  of  1  per  cent. 

A  50  * 

EXERCISE   XVIIL 

1.  In  the  illustrative  Example  1  of  Art.  107,  suppose  the  error  in 
measuring  the  edges  was  0.02  ft.,  0.03  ft.,  and  0.04  ft.,  respectively,  find 
the  approximate  error  in  the  volume  computed  with  6,  8  and  10  ft.  as 
the  edges.  Ans.  5.32  cu.  ft. 

(Exact  error,  AV  =  5.339624  cu.  ft.) 

2.  The  total  surface  of  a  cylinder  with  diameter  equal  to  altitude  is 
to  be  gilded  at  a  cost  of  10  cents  per  square  inch.  If  the  altitude  is 
measured  as  24  in.,  find  the  maximum  error  in  cost,  measurement  being 
accurate  to  ^  in. 

3.  The  period  of  a  pendulum  is  T  =  2tt  y  — .     Find  the  greatest 

error  in  the  period  if  there  is  an  error  of  dzxV  ft.  in  measuring  a  10  ft. 
L,  and  g,  taken  as  32  ft. /sec2,  may  be  in  error  ^  ft.  per  sec2.  Find 
the  percentage  error: 

Ans.  0.0204  sec,  ||  per  cent. 

4.  In  estimating  the  number  of  bricks  in  a  pile,  if  the  pile  is  measured 
to  be  8  X  50  X  5  ft.,  and  the  count  is  12  bricks  to  the  cubic  foot,  find  the 
cost  of  the  error  when  the  tape  is  stretched  2  per  cent  beyond  the 
standard  length,  bricks  being  sold  at  S10  per  thousand. 

5.  If  the  side  c  of  a  triangle  ABC  is  determined  by  measuring  the 
sides  a  and  6  and  the  included  angle  C,  show  that  the  error  Ac,  due  to 
inaccurate  measurements,  is  given  approximately  by  the  equation, 

Ac  =  Aa  cos  B  +  A6  cos  A  +  aAC  sin  B. 


164  DIFFERENTIAL  CALCULUS 

6.  If  the  horse  power  of  a  steamship  is  given  by  the  formula  H  = 
Kifi  D*,  show  that  the  increase  in  horse  power,  due  to  an  increase  Av 
in  the  speed  and  an  increase  AD  in  the  displacement,  is  given  approxi- 
mately by  the  equation, 

AH  =  SKifiD1  •  Av  +  %  KifiD-i  -  AD. 

7.  Show  that  the  relative  error  in  the  area  of  the  ellipse  due  to 

inaccurate  measurements  of  the  semi-axes  a  and  b  is  given  approxi- 

,   .     ,          AA       b'  Aa  +  a'  Ab 
matelyby     -j-  = ~b 

8.  The  equation  for  the  length  L  and  the  period  T  of  a  pendulum 
being  4  tt2L  =  T2g,  if  L  is  calculated  taking  T  =  \  and  g  =  32  ft. /sec2, 
while  the  true  values  are  T  =  1.02  and  g  =  32.01  ft. /sec2,  show  that 
the  approximate  error  in  L  is  AL  =  0.0326  .  .  ft.,  and  the  percentage 
error  about  4  per  cent. 

9.  In  determining  specific  gravity  by  the  formula  s  =  A/ A  —  W, 
where  A  is  the  weight  in  air  and  W  the  weight,  find  (a)  approximately 
the  maximum  error  in  s  if  A  can  be  read  within  0.01  lb.  and  W  to  0.02  lb., 
the  actual  readings  being  A  =  9  lb.,  W  =  5  lb.,  find  (6)  the  maximum 
relative  error. 

Ans.  (a)  As  =  0.0144; 
,..    As         23         23 
(6)  T  =  3000  =  36perCent- 

109.  Partial  Differentials  and  Derivatives  of  Higher 
Orders.  —  If  only  one  of  the  independent  variables  is 
supposed  to  vary  at  the  same  time,  by  successive  differen- 
tiations there  are  formed  the  successive  partial  differentials 
dx2u,  dy2u,  dx3u,  dy3u,  ...  or 

d2u  1  ,       d2u  7  „       d3u  .  ,       d3u  ,  , 

wdx'  dfdy>  a?^-  wd* 

For  example,  if  u  =  x2  +  xy2  +  y2,  (1) 

dxu  =  (2  x  +  y2)  dx,        dx2u  =  2  dx2,  dxsu  =  0; 

dyu  =  (2  xy  +  2  y)  dy,     d2u  =  (2  x  +  2)  dy2f     dyzu  =  0. 

If  u  is  differentiated  with  respect  to  x,  then  the  result  with 
respect  to  y,  there  is  gotten  the  second  partial  differential^ 

d™u  or  £rydxdy- 


INTERCHANGE  OF  ORDER  OF  DIFFERENTIATION       165 

For  example,  if  u  =  xz  +  x2y2,  (2) 

dxu  =  (3  x2  +  2  xy2)  dx,     dxy2u  =  4  xy  dx  dy. 

Similarly,  the  third  partial  differential  dvJu  or  2  d?/  da;2 

denotes  the  result  gotten  by  differentiating  u  once  with 
respect  to  y,  then  this  result  twice  successively  with  respect 
to  x. 

The  symbols  for  the  partial  derivatives  are: 

d2u         d2u         d2u       dzu         d3u 


»       ~; — r~ i      "r-^  '       ~r~?> 


dx2       dx  dy      dy2       dxz       dy  dx2 ' 

In  getting  the  successive  partial  differentials  and  deriva- 
tives of  u  or  /  (x,  y)}  dx  and  dy  are  treated  as  constants, 
since  x  and  y  are  independent  variables,  varying  by  uniform 
increments.  The  equivalent  symbols  for  the  higher  partial 
derivatives  by  another  notation  are  for/  (x,  y), 

/."(*,*),  Uy\x,y),  /„"(*, y),  h"\x,y),  f,*'"(x,y)t 

110.   Interchange  of  Order  of  Differentiation.  — 

r*  t(       \         &u  d2u  ,  . 

If  »-/fe»).     d^y^dtTx'  (1) 

d3it    _      d3w.      _    d3w 
da:2 d?/  ~  dxdydx~  dydx2}       '' 

that  is,  (f  w  is  differentiated  successively  m  times  with  respect 
to  x  and  n  times  with  respect  to  y,  the  result  is  independent  of 
the  order  of  these  differentiations. 

It  can  be  shown  that  the  order  is  always  a  matter  of 
indifference  if  fx'(x,  y),  fxy"{x,  y)  or  fy'(x,  y),  fyx"(x,  y)  are 
continuous  functions  of  the  two  variables  (x,  y)  taken 
together. 

In  most  cases  that  call  for  the  application  of  the  methods 
of  the  Calculus  to  physical  problems  the  partial  derivatives 
give  the  same  result  in  whatever  order  the  differention  is 
done. 


166  DIFFERENTIAL  CALCULUS 

For  example,  to  verify  the  theorem  in  some  cases: 
Example  1.  —  Given  u  =  ex  cos  y; 


du                                 d  1 

Tx  =  emsy'        dy\ 

Kdxj 

|           d2U 

=  -j — r-  =  —  ex  sin  1 
'      dydx 

du                  .               d  I 
-=-e*Smy,      ^( 

4yj 

I  -     dhi   -        e'sin 
'      dxdy 

mple  2.  —  Given  u  = 

iz. 

du      log  z 

d2U            1   . 

dx         y 

dz  dx      yz ' 

du       X 
dz      yz1 

d2u        1  . 

dx  dz      yz ' 

du          x  log  z 

d2u            log  z . 

dy              y2 

dx  dy            y2 

du      log  z 

d2u             log  z . 

dx        y 

dy  dx             y2 

du      x 

d2u              x  . 

dz      yz1 

dy  dz           y2z ' 

du          x  log  z 

d2w              x 

dy              y2 

dz  dy           y2z 

EXERCISE   XIX. 

Verify  the  identities  (1)  and  (2)  of  Art.  110  in  each  of  the  following 
nine  examples: 

1.    u  =  cos  (x  +  y).         2.   u  =  exsmy.  3.   u  =  cosxy2. 

4.   u  '=  xzy2  +  ay3.  5.   u  =  log  (x2  +  y2).     6.   u  =  ?/x. 

7.   u  =  xy  cos  (x  +  w).     8.   m  =  tan-1  -.  9.   u  =  sin2  x  cos  y. 

10.    llu  =  {x  +  y)\  *_  +  „_=_. 

11.  ifM  =  (x2  +  M  *g+9*aS+*3-a 

1  d2M       a2M        d2U  _ 

12,    Uu  =  (x2  +  2/2  -f-^i'  dx2  +  <fy2  +  dZ2   ~     ' 


EXACT  DIFFERENTIALS  167 

13'    Ifw=e*"  d^dz  =  {l+*XyZ  +  x2y2z2)u' 

14.    If  u  -  sin-  (xyz),  ■«*      =  i±^L. 

dxdydz       q  _  afys^)! 

111.   Exact  Differentials.  —  An  expression  of  the  form, 
Mdx  +  Ndy,  .(1) 

where  3/  and  A'  are  functions  of  x  and  y,  may  or  may  not  be 
the  differential  of  some  function  of  x  and  y;  if  it  is,  it  is  called 
an  exact  differential.  Some  simple  expressions  may  be  seen 
at  once  to  be  exact  differentials;  thus,  y  dx  +  xdy  is  an  exact 
differential,  for  it  is  recognized  as  the  total  differential  of 
xy. 

If  M  dx  +  N  dy  is  an  inexact  differential,  no  function 
F  (x,  y)  can  be  found  the  differentiation  of  which  will  give 
this  differential;   thus  y  dx  —  x  dy  is  an  inexact  differential. 

In  applying  the  Calculus  to  problems  in  physics  and 
mechanics  expressions  like  (1)  frequently  arise  and  some 
test  is  needed  to  determine  whether  the  expression  can  be 
gotten  by  the  differentiation  of  any  function  of  the  variables 
involved. 

As  an  example,  the  work  W  of  moving  a  particle  in  the 
XY  plane  gives  rise  to  the  expression 

dW  =  Xdx  +  Ydy,  (2) 

where  X  and  Y  are  respectively  the  x-  and  ^-components  of 
the  force  acting  on  the  particle.  Since  work  is  the  product 
of  force  by  distance, 

v      dW  a     v     dW 

X  =  -r—     and     Y  =  —r- » 
dx  dy 

and  (2)  takes  the  form 

iw      dTF  ,     .  dW  ,  ,Q. 

dW  =  ^dx  +  -!ydy'  (3; 

Here  (3)  was  not  gotten  by  differentiation  of  any  function 
W  =f(.x>y)   and  the  question  is  whether  it   could  be  so 


168  DIFFERENTIAL  CALCULUS 

gotten.     In  general,  if  M  and  N  are  any  chosen  functions  of 
x  and  y,  does  a  function  of  the  independent  variables  (x,  y) 
exist  that  will  upon  differentiation  give  Mdx  +  Ndy? 
If  there  is  such  a  function  u  =  F  (x,y),  then 

dw  =  Txdx  +  fydy-        I  (4) 

Now  if  the  differentiation  of  the  given  function  gives 

Mdx  +  Ndy,  (1) 

a  comparison  with  the  exact  differential  given  in  (4)  gives 

M  =  d~x'        N  =  Ty'  (5) 

that  is,  M  and  N  must  be  the  partial  derivatives  of  the 
function  u  with  respect  to  x  and  y,  respectively. 

According  to  the  theorem  of  Art.  110,  differentiating  (5) 

gives 

dM  =     d2u    =  5iV 

cfy       dx  dy      dx 
Hence,  if  M  =  -j-  and  JV  =  -r- ,  it  is  manifest  that 

2M  =  -^  (6) 

di/        dx 

is  the  necessary  condition  that  M  dx  +  N  dy  may  be  gotten 
by  the  differentiation  of  a  function  F  (x,  y),  and  it  may  be 
shown  that  it  is  a  sufficient  condition. 
•  When  the  condition  (6)  is  satisfied,  M  dx  +  JV  dy  is  an 
exact  differential;  when  the  condition  is  not  satisfied,  M  dx  + 
AT  di/  is  an  inexact  differential. 

Example  1.  —  Given  M  dx  +  N  dy  =  ydx  +  xdy. 

TT  IT  AT  dM  ,  dN  ^ 

Here       M  =  y,    N  =  x,     -j-  =1,     -j-  =  1. 

*  a?/  ax 

The  condition  (6)  is  satisfied  and  ydx  +  xdy  is  an  exact 
differential.    The  test  is  hardly  needed  in  this  simple  case, 


EXACT  DIFFERENTIALS  169 

as  it  may  be  seen  at  once  that  the  function  sought  is  xy+C, 
where  C  is  a  constant,  positive  or  negative,  or  zero. 
Example  2.  —  Given  M  dx  -\-  N  dy  =  y  dx  —  x  dy. 

Here,  since   -=—  =  1  and  -=—  =  —  1,  the  condition  (6)  is 

not  satisfied ;  hence,  y  dx  —  x  dy  is  an  inexact  differential, 
and  no  function  of  (x,  y)  exists,  the  differentiation  of  which 
will  give  this  differential . 

Note.  —  If  the  equation  ydx  —  xdy  =  0  is  given,  it  may 
be  changed  to  an  exact  differential  equation;  M  dx  +  N  dy  = 
0,  being  called  an  exact  differential  equation  when  M  dx  + 
N  dy  is  an  exact  differential. 

Thus,  multiplying  by  y~2,  the  equation  given  becomes 
ydx  -  xdy  _ 
V2 
which  is  exact,  and  the  function  F  (x,  y)  is  given  by  x/y  =  C. 
Again,  multiplied  by  1/xy,  the  equation  given  becomes 

*?  _  c!e  =  o 

x        y 

which  is  exact,  and  the  function  F  (x,  y)  is  given  by  log  x/y  = 
log  C:  Either  of  these  results  evidently  implies  the  other. 
Multiplying  the  equation  given  by  —x~2  gives  F  (x,  y)  by 

y/x  =  Ci. 

Example  3.  —  Given  Mdx  +  N  dy  =  -dx  +  log  x  dy. 

XJ  JLT        V        XT        1  BM         \  dN         1 

Here       M  =  -,     N  =  log  x,     -^  =  -,      —-  =  -■ 
x  dy       x       dx       x 

The  condition  (6)  is  satisfied  and  the  differential  is  exact. 
It  is  easy  to  recognize  that  F  (x,  y)  is  in  this  case  y  log  x. 
Example  4.  —  Given  M  dx  +  N  dy  =  sin  y  dx  +  x  cos  y  dy. 

Here  M  =  sin  y,    N  =  x  cos  y,    -=—  =  cos  y,     -j-  —  cos  y. 

The  condition  (6)  is  satisfied  and  the  differential  is  exact. 
The  function  F  {x,  y)  may  be  seen  to  be  x  sin  y. 


170  DIFFERENTIAL  CALCULUS 

Example  5.  —  Given  x  dy  —  y  dx.  Change  to  value  in 
polar  coordinates,  by  x  =  p  cos  0,  y  =  p  sin  0;  x  dy  —  y  dx  = 
p2  dd.    Dividing  by  x2, 

xdy-ydx=_ldB       ^ 

X2  p2  cos2  0 

where  the  differentials  are  exact,  and  the  function  F  (x,  y)  is 

y/x  =  tan  0.     (See  Note  Example  2.) 


EXERCISE   XX. 

Determine  which  of  the  following  differentials  are  exact,  and  for 
such  as  are  exact  find  the  functions  that  differentiated  would  give  them : 

1.  y  sin  2  x  dx  +  sin2  x  dy.  Ans.  y  sin2  x. 

2.  (ye*  +  ev)  dx  +  (ex  +  xev)  dy.  Ans.  yex  +  xey. 

3.  (y3  -  2  xy)  dx  4-  (3  xy2  -  x2)  dy.  Ans.  yzx  -  x2y. 

4.  vndp  +  npvn-ldv.  Ans.  pvn. 

5.  ex  sin  y  dx  +  e*  cos  y  dy.  Ans.  ex  sin  y. 

v2 

6.  —  da:  +  x  log  z  d?/.  Ans.  Inexact. 
x 

7.  (x2  —  y)dx  -  xdy.  Ans.  \  x3  —  xy. 

8.  ex  (x2  +  y2  +  2  x)  dx  +  2  ex2/di/.  Ar?s.  <?*  (.r2  +  ?/2). 

112.   Exact  Differential  Equations.  —  Equations  of  the 

form 

M  dx  +  N  dy  =  0, 

are  called  exact  differential  equations  when  M  dx  +  N  dy  is 
an  exact  differential,  the  total  differential  of  some  function 
of  (x,  y),  M  and  N  being  functions  of  x  and  y.  The  solving 
of  differential  equations  involves  the  Integral  Calculus,  and 
the  preceding  Article  with  the  Examples  and  Exercise  are 
introductory  to  the  subject. 

The  finding  of  the  function  from  which  an  exact  differen- 
tial may  be  gotten  by  differentiation  is  essentially  Integra- 
tion, the  inverse  process  to  Differentiation. 

In  Part  II  on  the  Integral  Calculus  the  subject  of  differen- 
tial equations  is  given  further  treatment. 


PART  II. 
INTEGRAL  CALCULUS. 


CHAPTER  I. 
INTEGRATION.  STANDARD  FORMS. 

113.  Inverse  of  Differentiation.  —  It  has  been  shown 
that,  when  a  function  is  given  and  its  rate  of  change  is 
required,  the  derivative,  which  expresses  the  rate  of  change, 
is  gotten  by  the  differentiation  of  the  function. 

It  often  occurs  that  the  rate  of  change  of  a  function  is 
known  and  the  value  of  the  function  is  desired.  In  many 
problems  in  pure  and  applied  mathematics  the  derivative  or 
the  differential  of  some  function  is  given  and  the  function 
itself  is  required. 

The  derivative  or  the  differential  of  a  function  being  given,  > 
it  is  a  natural  inference  that  an  inverse  operation  to  differen- 
tiation should  yield  the  function.  This  inverse  operation, 
the  opposite  of  differentiation,  is  called  integration  and  to 
integrate  any  given  function  (which  when  continuous  is 
always  the  derivative  of  some  other  function)  means  to  find 
that  other  tunction  whose  derivative  is  the  given  function. 
The  function  to  be  found  is  called  an  integral  of  the  given 
function,  which  is  called  the  integrand;  that  is,  a  function 
is  an  integral  of  its  differential.  The  process  of  finding  an 
integral  of  a  given  function  is  integration,  the  inverse  of 
differentiation;  that  is,  integration  is  anti-differentiation  and 
an  integral  is  an  anti-differential. 

171 


172  INTEGRAL  CALCULUS 

When  dy  =  d{f(x));  d~l  (dy)  =  d~l (df  (x)) ,  (read  "the 
anti-differential  of  dy  equals  the  anti-differential  of  d{J(x))"  is 
the  inverse  expression,  reducing  to  y  =  f  (x)}  as  the  two 
symbols  neutralize  each  other.     The  sign  of  integration  is, 

however,    /  ,  an  elongated  S;    and  this  symbol  indicates 

that  the  differential  expression  before  which  it  appears  is 
to  be  integrated,  the  whole  expression  denoting  the  integral 
itself. 

Thus    Cdy^d-^dy)     and      f d  (f (x))  =  d~l (df  (x)) ; 

the  sign  of  integration  and  the  symbol  of  differentiation 
indicating  inverse  operations  here  neutralize  each  other,  so 


/ 


d(f(x))=f(x). 

There  is  here  a  close  analogy  with  the  algebraic  signs  of 
evolution  and  involution;  for  example,  Vx2  =  x,  the  two 
symbols  indicating  inverse  operations  neutralizing  each 
other.  The  analogy  extends  further  to  the  fact  that,  while 
the  operation  of  raising  a  given  number  to  the  second  or 
other  power  is  a  direct  operation  and  involves  no  difficulty 
in  any  case,  the  inverse  operation  of  extracting  a  root  may 
not  be  done  so  directly  and  in  many  cases  can  be  done 
approximately  only.  While  it  has  been  shown  that  every 
continuous  function  has  an  integral,*  this  integral  may  not 
be  expressible  in  terms  of  the  elementary  functions.  In 
such  cases,  however,  an  approximate  expression  for  the 
integral  may  be  obtained  by  infinite  series  or  by  the  measure- 
ment of  an  area  representing  the  integral.  Most  of  the 
functions  that  occur  in  practice  can  be  integrated  in  terms 
of  elementary  functions,  either  directly  by  the  knowledge 
acquired  from  differentiation,  by  reversing  the  rules  of 
differentiation,  or  by  reference  to  a  table  of  integrals.    ' 

*  By  Picard,  in  Traite  d' Analyse. 


INDEFINITE  .INTEGRAL  173 

Except  for  simple  differential  expressions  the  process  of 
integration  is  less  simple  and  easy  than  the  process  of  differ- 
entiation. Just  as  any  finite  number  can  be  raised  to  a 
power,  so  can  any  finite  continuous  function  be  differentiated; 
and  as  the  roots  of  some  numbers  can  be  expressed  approxi- 
mately only,  so  the  integrals  of  some  functions  can  be 
expressed  approximately  only. 

There  is  one  function  whose  integral  is  not  some  other 
function  but  is  the  function  itself.     This  is  the  function  ex, 

whose  derivative  is  ex.    As   I  exdx  =  e 


I  exdx  =  ex,  so  Vl  =  1;    the 


particular  analogy  in  this  exceptional  case  is  manifest. 
114.   Indefinite  Integral.  —  When 

g  =  /'  (x)   or  dy  =  f  Or)  dx,  y  =  Jf  (x)  dx, 

read  "y  is  equal  to  an  integral  of  f  (x)  dx."  An  integral  of 
dy  is  evidently  y,  and  /  (x)  is  an  integral  of  its  differential 
f'(x)dx. 

Thus  integrals  of  many  simple  differential  expressions  are 
known  directly,  by  merely  recalling  the  function  which 
differentiated  results  in  the  given  expression. 

However,  since  the  differential  of  any  constant  term  of 
a  function  is  zero,  the  function  sought  may  contain  a  con- 
stant or  constants  no  indication  of  which  appears  in  the  given 
differential  or  derivative. 

Hence,  the  integral  of  a  differential  expression  is  in 
general  indefinite,  owing  to  the  lack  of  knowledge  as  to 
the  existence  or  value,  if  existent,  of  constant  terms  of  the 
function  sought. 

If  F  (x)   is  a  function  whose  derivative  is  /  (x),   then 


/ 


/  (x)  dx  =  F  (x)  +  C,  is  the  indefinite  integral,  where  C 

is  a  general  constant,  called  the  constant  of  integration,  de- 
noting a  value  either  positive  or  negative  or  zero. 


174 


INTEGRAL  CALCULUS 


When 


dy 
dx 


115.   Illustrative      Examples.  —  Example      1 

f  (x)  =  m,  is  given  as  the  constant  slope  of  y  =  /  (x) ; 

then  y  =    I  mdx  =  mx  +  C.    The  result  is  indefinite  be- 

m,  is  the  slope  of  y  =  mx  +  C,  y  =  mx 


cause 


dy 
dx 


Cor 


y  =  mx.  The  constant  C  added  to  right  member  of  the 
equation  includes  all  constant  terms,  if  any,  of  the  function; 
for  if  the  result  be  written  y  +  C"  =  mx  +  C",  then  y  = 
mx  +  C"  -C  =  mx+C.  The  letter  C,  often  omitted, 
should  be  written  as  part  of  the  result  of  the  integration. 

Data  may  be  available  in  some  cases  to  make  the  value  of 
C  known,  or  to  eliminate  it,  and  thus  to  make  the  result 
determinate. 

In  this  example,  if  it  is  known  that  the  function  has  the 
value  b  when  x  is  zero,  then  y  =  mx  +  b,  since  C  is  equal  to 
b  when  x  is  zero.  If  y  is  —b,  or  if  y  =  0,  when  x  =  0;  then 
y  =  mx  —  6,  or  y  =  mx. 


As  shown  in  the  figure  the  function  is  a  straight  line 
making  an  angle  0  ( =  tan-1  m)  with  the  X-axis,  the  con- 
stant of  integration  being  the  F-intercept.  The  indefinite 
or  general  integral  is  y  =  mx  +  C,  any  straight  line  with 
slope  m. 


ILLUSTRATIVE  EXAMPLES 


175 


Note.  —  When  the  value  of  C  is  determined  the  integral 
is  called  a  particular  integral. 

Example.  2.  —  When   -~  =  2  x  is  given  as  the  rate  of 

change  of  y  with  respect  to  x,  then  y  =    I  2  x  dx  =  x2  +  C, 

where  x-  +  C  is  the  general  integral  of  2  x  dx,  since  the 
differential  of  (x2  -f  C)  is  2  a;  efo.  Here  z2  +  C  is  a  function 
whose  rate  of  change  is  2  x,  and 
if  the  rate  of  change  is  that  of 
the  ordinate  to  the  abscissa,  or 
the  slope  of  a  curve,  then  the 
integral,  y  =  x2  +  C,  is  the  equa- 
tion of  the  curve.  The  locus  of 
the  equation  is  a  parabola  with 
its  vertex  at  a  distance  C  above 
or  below  the  origin,  or  at  the 
origin,  according  as  the  value 
of  C  is  positive,  negative,  or 
zero. 

If  y  is  known  for  some  value  of  x,  then  the  value  of  C  is 
easily  determined.  For  instance,  if  it  is  known  that  the 
point  (a,  6)  lies  on  the  curve,  then  y  =  x2  +  C,  must  be  satis- 
fied by  the  coordinates  (a,  b),  giving  b  =  a-  +  C,  and,  there- 
fore, C  —  b  —  a2.  Hence,  the  particular  parabola  is  y  =  x2  + 
b  —  a2.  If  the  curve  is  known  to  pass  through  the  origin, 
then,  since  C  is  zero,  y  =  x2  is  the  parabola  with  vertex  at 
the  origin. 

dA 
Example  3.  —  If  a  given  derivative  -r-  =  2  x  represents 

the  rate  of  change  of  an  area  A  to  a  length  x,  then  A  = 

j  2  x  dx  =  x2  +  C,  is  an  area  where  x2  may  represent  the 

area  of  a  variable  triangle  formed  by  the  straight  line  y  =  2  x, 
the  ordinate  at  any  value  of  x,  and  X-axis.     In  the  figure 


176 


INTEGRAL  CALCULUS 


shown  the  area  A  is  zero  when  x  is  zero,  and  therefore  C  is 
zero.  The  area  of  any  triangle  being  one-half  the  product 
of  base  and  altitude,  the  result  of  the  integration,  A  —  x2, 
is  seen  to  be  true.  The  area  A  =  x2  +  C  may  represent  a 
square  of  side  x  and  some  additional  area  represented  by  C, 
undetermined,  as  it  might  be  positive,  negative,  or  zero  — ■ 
the  derivative  of  the  area  in  either  case  being  the  given 
rate  2  x. 


dA 
Example  4.  —  If  the  given   derivative  is  -p-  =  x2,  then 


-/ 


XA 


x2  dx  =  -=  +  C,    since 


d(f+c)  = 


x2dx. 


Here  the  area  ■=■  is  that  bounded  by  the  curve,  a  parabola 

y  =  x2,  the  ordinate  at  any  value  of  x,  and  the  X-axis.*  In 
the  figure  shown  the  area  A  is  zero  when  x  is  zero,  and  there- 
fore C  is  zero. 

It  is  seen  that  the  area  OPM  is  exactly  one-third  of  the 
area  of  the  circumscribed  rectangle.  Hence,  the  area  OPN 
between  the  curve,  the  abscissa  at  the  end  of  any  ordinate, 
and  the  F-axis,  is  two-thirds  the  area  of  the  same  rectangle. 


ILLUSTRATIVE  EXAMPLES  177 

Example  5.  —  The  acceleration  of  a  falling  body  being 

nearly  constant  near  the  earth's  surface,  it  is  required  to 

find  the  velocity  and  the  distance  after  any  time.     If  s 

denotes  the  distance  along  a  straight  line  positive  upward 

d2s 

and  t  the  time,  then  -==  is  the  acceleration. 

at2 


dv      d2s 
di 


a<s  _     \dtj  /cis\ 

w~ir=-^  or  d\dt)  =  ~gdL     (1) 


Integrating  gives  velocity, 

v  =  ~=-gt  +  (C  =  v0),  (2) 

where  v0  is  the  initial  velocity,  when  t  =  0;  then, 
ds=  —gtdt-\-  Vq  dt. 
Integrating  gives  distance, 

s=  -i$2+M  +  (e  =  so),  (3) 

where  s0  is  the  initial  distance,  when  t  =  0.  If  the  body  falls 
from  rest,  v0  and  sQ  are  zero,  hence;  v  =  —gt;  s  =  —\gt2', 
and  v2  =  —  2  gs,  by  eliminating  t.     (See  Art.  14.) 

Example  6.  —  Determine  v  and  s  in  terms  of  t  for  a  bullet 
shot  vertically  upward  with  a  velocity  of  2000  feet  per 
second,  neglecting  air  resistance. 

dv      d2s  . 

~Tf  —  ~JE=  —32.2  it.  per  sec.  per  sec. 

V  =  It=JlJ=J  -S2-2dt=  -32.2*  +(C=v0'=  2000), 

u  =  0o>     when  t  =  0. 
8=   Jds=  j  -32.2 tdt  + 2000 dt=  -lQ.lt2 +2000 1 

+  (C  =  so  =  0),     s  =  so  =  0,     when  t  =  0. 
To  find  the  time  of  rising,  make  v  =  0  =  -32.2  t  +  2000; 
/.     t  =  62.1  sec. 


178  INTEGRAL  CALCULUS 

To  find  the  height  i\  will  rise,  s  =  -16.1(62.1)2  +  2000 
(62.1)  =  62,112  ft. 
To  find  the  time  of  flight,   s  =  0  =  -16.1  t2  +  2000  t; 

:.     t  =  124.2  sec.     and     t  =  0. 

Hence,  the  time  of  falling  is  the  same  as  that  of  rising, 
since  the  time  of  flight  is  twice  that  of  rising.  The  height  it 
will  rise  may  be  found,  by  making  v  =  0  in 

2        2      o„0.      •      c       v*       (20QQ)2      R9  no* 
vz  =  v02-2gs;     ..     s  =  ^-  =     64  4     =  62,112 It., 

the  same  as  above. 

Remarks.  —  These  examples  illustrate  the  important  fact 
that  the  knowledge  of  the  rate  of  change  of  a  quantity  to- 
gether with  the  knowledge  of  its  original  value,  makes 
possible  the  complete  determination  of  the  value  of  that 
quantity  at  any  time.  This  must  be  so,  since  two  different 
quantities  with  the  same  rate  of  change  always  have  a  con- 
stant difference,  the  rate  of  change  of  their  difference  being 
zero.  This  is  in  accordance  with  the  undoubted  fact  that 
if  the  rate  of  change  of  a  quantity  decreases  to  zero  and 
remains  zero,  the  quantity  ceases  to  change  at  all,  being 
then  constant.  The  fact  is  formulated  in  principle  (iv)  of 
Art.  116,  and  is  the  converse  of  the  fact  that  if  a  quantity  is 
constant,  its  rate  is  zero. 

116.  Elementary  Principles.  —  While  there  is  a  general 
method  of  differentiation,  for  the  inverse  process  of  integra- 
tion no  general  method  has  been  devised.  For  the  integra- 
tion of  the  various  differential  expressions,  rules  have  been 
formulated  and  special  methods  have  been  found,  one  or 
more  of  which  provide  for  every  case  in  which  integration 
is  possible. 

These  rules  or  formulas  are  derived  or  disclosed  through 
knowledge  of  the  rules  of  differentiation;  in  fact,  the  rules 
most  used  are  merely  directions  for  retracing  the  steps  taken 
in  differentiation. 


ELEMENTARY  PRINCIPLES  179 

Elementary  principles  that  apply  in  integration  may  be 
expressed  as  follows: 


(i) 


Jf(x)dx^F(x)+C,     if     dF(x)=f(x)dx. 


—      =  log  x  +  C,        since    d  (log  x)  =  — 
x  x 


This  principle  furnishes  the  most  direct  proof  of  formulas 
for  indefinite  integration,  and  provides  a  decisive  test  of  the 
correctness  of  the  result  of  any  integration.     Thus, 

lxndx  = — — r  -f-  C,      since     d\ — —)  =  xndx\ 
J  n  +  1  \n-\-lJ 

i 

/Ifdx  == — r  +  C,        since    dU — A=bxdx. 
log  b  \\og  bj 

In  this  manner  the  test  can  be  applied  to  prove  any  formula, 
or  to  verify  the  result  of  the  integration  of  any  expression. 

(ii)  A  constant  factor  can  be  transposed  from  one  side  of  the 
sign  of  integration  to  the  other,  and  a  constant  factor  can  be 
introduced  on  one  side,  if  its  reciprocal  is  introduced  on  the 
other,  without  changing  the  value  of  the  integral. 

For,  if  a  is  a  constant, 


I  ay  dx  =  a   I  y  dx, 


since  the  differentiation  of  the  equation  gives,  ay  dx  =  ay  dx. 
Hence, 

fydx  =  \faydx  =  af\ydx. 

(iii)    The  integral  of  a  polynomial  is  equal  to  the  sum  of  the 
integrals  of  its  several  terms.    For 

/  (<£+  x  —  x2)  dx  =    /  adx  +   I  xdx  —   I  x2dx, 

since  the  differentiation  of  the  equation  gives 

a  dx  +  x  dx  —  x2  dx  =  a  dx  +  x  dx  —  x2  dx. 


180  INTEGRAL  CALCULUS 

(iv)     fo  =  C,    since    dC  =  0. 

The  integral  of  zero  is  a  constant. 

ds 
Thus,      if  -=7  =  v,  where  v  is  constant  velocity, 

d^s      dv 

-Tp  =  -r  =  0;  that  is,  acceleration  is  zero, 


hence 


.  js- /.-«-. 


117.  Standard  Forms  and  Formulas.  —  There  follows  a 
list  of  standard  integrable  forms,  that  is,  differential  functions 
whose  integrals  can  be  expressed  in  finite  forms  involving 
no  other  than  algebraic,  trigonometric,  inverse  trigonometric, 
exponential,  or  logarithmic  functions.  To  integrate  a  func- 
tion that  is  not  expressed  in  terms  of  an  immediately  inte- 
grable form,  it  is  reduced  if  possible  to  one  or  more  of  such 
forms  and  the  formula  applied. 

The  formulas  in  general  are  gotten  by  merely  reversing 
theiormulas  for  differentiation,  and  each  can  be  proved  by 
the  principle  (i)  of  Art.  116. 

The  list  will  be  found  to  contain  the  one  or  more  than  one 
integral  to  which  every  integrable  form  is  reducible.  These 
forms  may,  therefore,  be  called  fundamental  although  only 
the  first  three  are  really  fundamental,  since  each  of  the  others 
by  substitutions  can  be  reduced  to  one  of  the  three.  While 
the  list  is  of  standard  integrable  forms,  it  may  be  supple- 
mented by  other  integrable  forms;  but  no  list  of  forms  is 
exhaustive,  even  when  extended  into  tables  of  integrals. 
After  the  acquirement  of  familiarity  with  the  rules  of  differ- 
entiation, and  the  common  methods  of  reduction  with  the 
standard  integrals,  the  use  of  tables  of  integrals  for  the  com- 
plicated forms  is  recommended;  much  time  otherwise  given 
to  formal  work  in  integration  being  thereby  saved. 


STANDARD  FORMS  AND  FORMULAS  181 

/xn+l 
xndx  =  — j-rr  +  C,  where  n  is  not  —1. 
tt  +  1 

/dx 
—  =  log  x  +  C  =  log  x  +  log  c'  =  log  (c'x). 

til.     fb*dx  =  r^-1-  +  C. 
J  log  6 

IV.  f(?dx  =  ex  +  C 

V.  I  sin  z  dx  =  —  cos  x  +  C,  or  vers  x  +  C. 

VI.  /  cos  x  dx  =  sinx  +  C,  or  —  covers  a;  +  C. 

VII.  /  sec2  xdx  =  tanx  +  C. 

VIII.  /  esc-  x  dx  =  —  cot  x  +  C. 

IX.  I  sec  x  tan  x  dx  =  sec  x  +  C. 

X.  /  esc  x  cot  x  dx  =  —  esc  z  +  C. 

XL  /  tan  z  dx  =  log  sec  x  +  C  =  —  log  cos  x  +  C. 

XII.  /  cot  xdx  =  log  sin  x  +  C  =  —log  esc  x  +  C. 

XIII.  /  esc  xdx  =  log  tan  ■=  +  C  =  log  (esc  x  —  cot  x)  -f  C. 

XIV.  J\ecs(fc  =  logtan(|  +  |)  +  C 

=  log  (sec  x  +  tan  x)  +  C 

XV.     f-r^  =  -tan-1-  +  C,     or    --cof^  +  C. 
J  a2  +  x2      a  a  a  a 

XVI.     f-^0  =  ^log^+C  =  -tanh-1-+C'.  (x2<a2) 
J  a2  —  x2     2a       a  —  x  a  a 

=  -^log— +C  =  -coth-1-+C'.  (x2>a2) 
2a    °jc— a  a  a 


182  INTEGRAL  CALCULUS 

XVII.     r^  =  ^log^+C=-icoth-^+C'.  (x2>a2) 
J  x2—a2    2a       x+a  a  a 

=  2-a^S+C=4tanh-1+C-  <*<*> 
XVIII.   f^=2  =  ^l  +  C,   or    -«ri5  +  (T. 

XIX      f    7  ^        =  log(x+V^+^)+C,orsinh-1--f-C'. 
J   vf  +  a2  a 

XX      f    /  dX       =  log(z+ViW)+C,orcosh-1--r-C'. 
J  Vx2  —  a2  a 

XXI      / — 7=  =  -  sec-1  -  +  C,  or   — csc^-H-C. 
J  x  Vx2  -  a2      a  a  a  a 

XXII.     f  a/o    *       2  =  vers"1  ?  +  C,  or  -  covers"1  ^  +  C". 
J   V2ax  —  x2  a  a 

The  two  forms  of  the  integral  in  several  of  the  formulas 
correspond  to  different  values  of  the  constants  denoted  by 
C  and  C.     Thus  in  formula  V, 

vers  x  -f-  cos  x  =  1  =  C  —  C", 
and  similarly  in  XVIII, 

sin-1-  +  cos-1-  =  ^=C,-C. 
a  a      2 

When  the  differentials  of  two  functions  are  equal,  their 
rates  are  equal;  therefore,  the  functions  will  be  equal  or 
differ  by  a  constant;  hence  the  variable  parts  of  the  indefi- 
nite integrals  of  the  same  or  equal  differentials  are  equal  or 
differ  by  a  constant. 

118.  Use  of  Standard  Formulas.  —  When  a  given  func- 
tion to  be  integrated  is  not  expressed  in  a  form  immediately 
integrable,  by  various  algebraic  and  trigonometric  trans- 
formations or  substitutions,  the  effort  is  made  to  reduce  it  to 
one  or  more  of  the  standard  forms  so  as  to  apply  the  formula. 
When  different  methods  are  used  the  results  may  have 


USE  OF  STANDARD  FORMULAS  183 

different  forms,  but  upon  reduction  they  will  always  be  found 
to  differ  (if  at  all)  only  by  a  constant,  in  accordance  with  the 
statement  above. 

Formula  I  is  the  standard  formula  for  the  Power  Form. 
It  is  of  most  frequent  application,  and  it  may  be  expressed 
in  words  as  follows : 

The  integral  of  the  product  of  a  variable  base  with  any  con- 
stant exponent  (except  —  1)  and  the  differential  of  the  base  is 
the  base,  with  its  exponent  increased  by  1,  divided  by  the  in- 
creased exponent,  and  a  constant. 

The  proof  has  been  given  in  (i),  Art.  116;  it  may  be  derived 
thus:  since 

f2xdx  =  x2  +  C,       I  3x2dx  =  x3  +  C,  etc., 

(ft  +  \)xndx  =  xn+l  +  C; 
hence,  in  general, 


/< 


/ 


/v.n+1 

xndx  =  — r-r  +  C- 
ft  -+-  1 


When  a  given  integrand  is  a  fraction  with  denominator 
to  a  power,  it  may  become  this  form  by  bringing  up  the 
variable  quantity  with  change  of  exponent's  sign;  but,  since 
the  variable  quantity  is  represented  by  x  in  the  formula,  it 
is  essential  that  the  differential  of  the  variable  quantity,  and 
not  merely  dx,  be  present  in  the  integrand  before  the  formula 
is  applicable. 

If  a  constant  factor  is  lacking,  it  may  be  supplied  in  ac- 
cordance with  (ii),  Art.  116;  but  it  should  be  noted  that  the 
principle  is  only  for  constant  factors. 

The  value  of  an  integral  is  changed  when  a  variable  factor 

is  transferred  from  one  side  of  the  sign   /  to  the  other;  thus, 
Cx2  dx  =  i  xs  +  C,     but     x   I  x  dx  =  |  x3  +  C. 


184  INTEGRAL  CALCULUS 

When  a  change  of  sign  is  needed,  the  constant  factor  —  1 
effects  the  change.     Thus,  for  an  example, 

r     xdx       =_1   f(a2_x2yh(-2xdx)=-V~tf^  +  C. 
J  v  a2  —  x2         ^  J 

To  verify : 

fd  (-Vo2-a^  +  C)  =    f-  \  (a2  -  x2)-*  (-2 xdx) 


i 


xdx 

■. ,  as  given. 


Va2  —  x1 

When  a  variable  factor  is  lacking,  resort  may  be  had  to 
expansion  and  then  application  of  the  formula  to  each  term 
of  the  polynomial.     Thus,  for  an  example: 

f2(l+x2)2dx  =  2  f(l  +  2x2+x*)dx  =  2(x+ix*+%xb)+C. 

When  the  numerator  is  of  higher  power  than  denominator, 
reduce  by  division  and  then  apply  formula  or  formulas,  thus : 

=  ix2  +  x  +  2\og(x-l)+C. 
If  n  =  —  1,  formula  I  gives  a  result  that  is  not  finite;  but 
when  n  =  —  1,  the  form  reduces  to  form  II  and  that  formula 
applies.     Thus, 

I  x~l  dx  =    /  —  =  log  x  +  C. 

Formula  II  may  be  stated  in  words  as  follows: 
The  integral  of  a  fraction  whose  numerator  is  the  differential 
of  its  denominator  is  the  Napierian  logarithm  of  the  denomina- 
tor, and  a  constant.     The  result  will  be  real  only  when  x  is 
positive.     When 

X  >  a>    J  x~^a  =  l0g  (x  ~  a)  +  C; 
but,  if 

x  <  a,      /  — —  =    /  ■ *  =  log  (a  —  x)+  C. 

J  x-  a      J  a-  x         &v  J 


USE  OF  STANDARD  FORMULAS  185 

Formula  III  may  be  stated  in  words  as  follows: 

The  integral  of  the  product  of  a  constant  base  with  a  variable 
exponent  and  the  differential  of  the  exponent  is  the  base,  with 
exponent  unchanged,  divided  by  the  Napierian  logarithm  of  the 
base,  and  a  constant. 

Here  the  base  b  must  be  positive  and  not  unity. 

Formula  77  is  the  special  case  of  III,  the  Napierian 
logarithm  of  base  e  being  unity. 

In  applying  these  two  formulas  to  given  integrands,  it 
is  essential  that  the  differential  of  the  variable  quantity, 
and  not  merely  dx,  be  present.     Thus,  for  examples : 

/***  =  wSh"^h^dX  =  I&6  +  C- 
I  ex'n  dx  =  n   I  ex'n  —  =  nex'n+  C. 

C^*dx  =  ^—    Cir^\0gTT^dx  =  ^-  +  C. 

J  3  log  7T  J  3  log  7T 

The  following  are  examples  of  integration  by  one  or  more 
of  the  first  four  standard  formulas. 

EXERCISE   XXI. 
In  these  examples  the  results  may  be  verified  by  (i),  Art.  116,  and 
the  verification  should  be  made  where  the  result  is  not  given. 

2.  f(ax  +  b)ndx  =  -  C(ax  +  b)nadx  =  {aX. +  ,6) **  +  C. 
J  a  J  a  (/!  t  1) 

3.  C(2a  +  3bxydx.  4.     C     ^dx    a. 
J  J  (a2  +  z3)* 

6.  jYi+^Ydx.  6.  fr-v-tzydt. 

7.  /2»*(£  +  l)  dV-  8-    f^2Y+lds. 

9.     C  \r2~pxdx  =  V2p  Cx*  dx  =  \  vTpz*  (=  f  x  V2px)  +  C. 
10.     f  {axn+b)pxn^dx  =  —  C  (axn+b)pnaxn-ldx  =  ^^r^+C. 


186  INTEGRAL  CALCULUS 


1.  f5xVl-2x*dx=-$j'a-2&)t(-±xdx)=-$(l-2a*)*+C. 

2.  jVT^Jxexdx=  -  f(l-ex)H-exdx)  =  -  §  (1  -  ex)%  +  C. 

„      /"adx  f  -nj         axi~n    .  /^ 

3.  I  — =-  =  a  I  x  ndx  =  ^ h  C. 

J   xn  J  1  —  n 

/£  a/vsa- 


a:2)  dx. 


6-  iVS? dx  "  3  JVt5? d*  -  a log  (a  +  te">  +  c- 

/•  -  (2  qg  -  x2)  dx  _  -  (3  ax2  -  x3)1 
J     (Sax* -x*)*  2 

22.    /^tto-.-f  +  f-logCr  +  D  +  Cf. 

23'    /ifc -I* <**>+* 

24.     f!?*"1"?  dx  =  x  +  log  (3x  -  1)§  +  C. 
./  o  x  —  1 

or      f        dx  r  e~x  dx 

J  ex(3+e-x)  ~  J  3+e~x' 

nn      C    sin  2  x     ,                                               __      /*    cot  x     , 
26.     I  — — r-^-  dx.  27.     (  i -. —  dx. 

J  1  +  sin2  x  J  log  sin  x 

28.    f{ex-e-x¥dx=  ^X~^  -2x+C.      29.    JV  +  e~«J  dx. 

30-  J?fif  -  /(-& + StS)  =  2l0g (e* + 2) " * + a 

-j^r-rdx.  32.     fa«fe*(te  =  1 ° .      ,  +  C. 

ez  +  1  J  log  a  +  log  6 

33.     f£»|*<te.  34.     f-^<fc. 

J  sm  2  x  »/  1  _|_  ^5 

«_      /*      log  x  dx  _  1    /*   —  21ogx  dx 

J  x(l  -log2x)  ~  ~  2  J   1  -  log2x~x' 
dx  /•  d.r/1  +  x2 


.     r  dx =  r 

J  (1  -f-  x2)  arc  tan  x      J 


tan-1  x 


DERIVATION  OF  FORMULAS  XI,  XII,  XIII,  AND  XIV       187 

119.   Derivation  of  Formulas  XI,  XII,  XIII,  and  XIV.  — 

By  the  application  of  Formula  II  the  following  results: 

r,tan  x  sec  x  dx 


r  r 

I  tan  xdx  =    J 


secx 
=  log  sec  x  +  C  =  -log  cos  a;  +C.  (XI) 

/cot xdx  =    I  - —  dx 
J  smx 

=  log  sin  x  +  C  =  -  log  esc  x  +  C.  (XII) 

/CSC  Xdx    =      /  -: 
J  smx 

dx 


-r- 


2  sin  x/2  cos  x/2 

S6Ctanl/t2  =  l0g  tan  X'2  +  C'      (XIH) 
Or 

fcscxdx  =    fcscx(-cotx  +  cscx)dx 
J  J  esc  x  —  cot  X 

=  log  (esc  x  —  cot  x)  +  C. 

I  sec  xdx  =    I  esc  (x  +  tt/2)  dx 

=  log  tan  (x/2  +  x/4)  +  C.  (XIV) 

Or 

f  secxtfz  =    pecs  (tan s  + sec g)fe 
J  J  sec  x  +  tan  a; 

'sec  x  tan  x  dx  +  sec2  a;  dx 


=/? 


sec  x  +  tan  x 
=  log  (sec  x  +  tan  x)  +  C. 

EXERCISE   XXH. 
By  Art.  116,  and  one  or  more  of  the  standard  formulas  I  to  XIV. 

1.  /(Bin3*+coBS*-Bin|)dx=-S2^+2^+2«»|+C. 

n      C  •  7         sin2  x   ,    ~ 

2.  I  sin  x  cos  xdx  =  — - 1-  C. 

o       C  -j  cos2  X    ,     _, 

3.  J  cos  x  sin  x  dx  = - 1-  C. 


188  INTEGRAL  CALCULUS 

4-  /(SS  +  aK)*-  -/«»r.»(-«in»d»)  +  /8in-.coe.*. 

J  smd  J        sin0  J 

6.  jsitfddd  =  j^—^^djd  =  \d-  iSin(20)+C. 

7.  /cos2^^/1  +  CO2S(20)^  =  ^  +  ^in(20)+C. 

8.  fsin30d0  =    f  (1  -  cos2  0)  sin  0  d0  =  -cos0  +  ^cos30  +  C. 

9.  Tsui3  0  cos3  0  d0  =    f  sin3  0  (1  -sin2  0)  d  sin  0  =  |  sin4  0-  §  sin6  0+C. 

0.  f  tan2  x  dx  =  J  (sec2  x  —  1)  dx  =  tan  a;  —  x  +  C. 

1.  f  cot2  xdx  =   I  (esc2  x  —  1)  dx  =  —cot  x  —  a;  +  C. 

2.  ftan3  0  de  =   ftan  0  (sec2  0  -  1)  d0  =  |  tan2  0  -  log  sec  0  +  C 
=  \  tan2  0  +  log  cos  0  +  C. 

3.  f  sec2  (ax2)  xdx  =  5-  tan  (ax2)  +  C. 

4-  /§S^=^^(a*)tan(o*)d(o*)= ^sec^+c- 

6.     (V*E__    f^#i=    fcsc(2z)d(2*) 
J  sin  x  cos  x      J  sm  (2  x)       J 

=  log  tan  x  +  C,  by  XIII. 

6.  ( =   (cscxsecxdx  =  f- dx  =  logtanx+C, by  II. 

J  sin  x  cos  x     J  J  tan  x 

7.  f  sec2  0  esc2  ddO  =   C  (sec2  0  +  esc2  0)  d0  =  tan  0  —  cot  0  +  C. 

20.      f^inxcoga,da.  21>     JVco9xsina.d;r 

22.    JVan(ax)sec2(ax)dx. 

120.  Derivation  of  Formulas  XVI,  XVII,  XIX,  and  XX.  - 

By  the  application  of  Formula  II,  the  following  results: 


DERIVATION  OF  FORMULAS  XVI,  XVII,  XIX,  AND  XX   189 
For  XVI,  put 


—  x2      2  a  \a  +  x 


a  —  Xj 

dx 


r   dx       j_  r  dx      j^  rz 

J  a2  —  x2      2aja  +  x      2a  J  a 


=  ^log(a  +  x)  _  _Llog(a _ x)  +  c 

=  s-  log h  C 

2a       a  —  x 

=  -  tanh"1  -  +  C".     (x2  <  a2)  (XVI) 

a  & 


or 


r  dx       r  -dx  =  i   ^  <fa      jl_  r  ^ 

J  a2  —  x2      J  x2  —  a2      2ajx-\-a      2a J  x  —  a 

=  J-log^±^  +  C  =  icoth-1  -  +  C".      (x2  >  a2) 

2a    *  x  —  a  a  a  y  ' 

The  first  or  second  of  these  results  is  used  according  as 
a  —  x  or  x  —  a  is  positive;    that  is,  the  form  of  the  result 
which  is  real  is  to  be  taken. 
For  XVII,  put 

1      =   1  /    1 LA 

x2  —  a2      2  a  \x  —  a      x  +  a  J 

r    dx       j_  r  dx        l   r  dx 

J  x2  —  a2      2ajx  —  a      2a J  vz  +  a 

=  2^  log  (x-  a)  -  —  log  (x  +  a)  +  C 
1  ,      x  —  a      n 

=  -  -  coth"1  -  +  C,     (x-  >  a2)       (XVII) 

a  a  /       v  / 


or 


r   dx        r  -dx  =  i    r  -dx      1    r 

J  x2  —  a2      J  a2  —  x2      2a  J  a  —  x      2a  J  a 


dx 


+  x 

J-log^^  +  C=  -itanh-^+C.    (x2  <  a2) 
2a    &a  +  £  a  a  v  J 


190  INTEGRAL  CALCULUS 

The  first  or  second  of  these  results  is  used  according  as  x  —  a 
or  a  —  x  is  positive. 
For  XIX,  let 

Vx2  +  a2  =  z  —  x;    or    z  =  x  +  Vx2  +  a2,  (1) 

/.     a2  =  z2  —  2  xz. 
d(a2)  =  0  =  2zdz  -  2zdz-  2zcfc; 
(2  —  x)  dz  =  zdx; 

dz        dx  dx  t      /-.x 

—  = =     ,         — ,     by  (1) 

2       z-x      Vx2  +  a2 

J  Vx2  +  a2     «/    « 

or  sinh"1  -  +  C".  (XIX) 

a 


For  XX,  similarly,  on  letting  Vx2  —  a2  =  z  —  x, 


i 


dx 


_  =  logGr+\/z2-a2)  +  C,  or  cosh"1-  +  C";  (XX) 
Vx2  —  a2  a 

The  logarithmic  form  of  cosh-1  -  is  log  ( x  "*"      x  ~  a  ) , 

but  its  derivative  or  differential  is  the  same  as  that  of  log 
(x  +  Vx2  —  a2),  the  constant  a  disappearing  in  the  differ- 

x 
entiation;  and  so  too  with  the  sinh-1-.     (See  Art.  66.) 

a 

121.   Derivation  of  Formulas  XV,  XVIII,  XXI,  and  XXII. 

—  These  formulas  are  merely  the  reverse  of  the  differential 
forms  given  in  Examples  1  and  2,  Exercise  VI. 

They  may  be  derived  from  the  forms  for  the  inverse  trigo- 
nometric functions  of  x .     Thus: 

r  dx 

+  x2      a    I    ,    .  x2      a    I    ,    .    fx\2      a  a 


since 


DERIVATION  OF  FORMULAS  XV,  XVIII,  XXI,  AND  XXII  191 

Since 

tan-1  -  =  I  -  cot"1  -,     d  (tan-1  -)  =  d(-  cot"1  -V 
a      2  a'       \  a/        \  a) 


Hence 


/ 


C  t2  =  -  tan-1  -  +  C,    or     --  cot"1  -  +  C.      (XV) 


a2  +  x2 

In  the  same  way  the  second  forms  follow  for  formulas  XVIII, 
XXI,  and  XXII. 

The  standard  forms  are  given  in  terms  of  -,  because  they 

are  of  more  use  than  those  in  terms  of  x;  the  latter,  being 
special  cases  where  a  =  1,  are  often  given  as  the  standard 
forms.  Integrals  may  be  obtained  by  reduction  to  either 
form. 


EXERCISE   XXIII. 

~b 


J  62  +  c2x2      c~Jb>  +  (ex)2  ~  be  b  +  U 

AJft*-c»j«      cJ62-(cx)2      26c10g6-cx  +  L     tcz  <&) 

^tanh-^  +  C 

OC  0 

cJ(cz)2-o2      2ocgcz-o  +  C     lCX>0j 

oc  6 

3  f  ^  f__^_  J_4     -t^  +  3      r 
*    Jx2  +  6^  +  12       J(x  +  3)2  +  3       V3  V3    + 

4  f         d:c  f  J*  _  1  lntr  (*  +  3)  -  2 

'   J  x*  +  6  x  +  5      J  (z  +  3)2  -  4      4     g  (x  +  3)  +  2  "l" 

1,      z  +  1    .    ~ 
=  -.  log  — —^  +  C. 

-      r  x2dx       1 ,      a;3  —  1   .    _  .      r     dx 


„      r   xdx  1  a:2       _  a      C       dx 

7-    jF+^  =  2^tan     ^  +  C-  8-    Jl6*T 


192  INTEGRAL  CALCULUS 

9-  /s^p  -  Ac  ><*  13  +c  -  -  £ coth- " +c-  ^  > 62) 


10 


?nr log  ^-^  +<?=  -  r tanh_1  t  +<?'•    (c2*2 <  &2) 
2  6c    &6+cx  be  b 

dx  n  r  2adx 


C  dx  =  2  f 

'    J  ax2  +  bx  +  c         J 


(2  ax  +  bY  +  4  ac  -  62 

tan  *  +  C     (4  ac  >  62) 


v  4  ac  -  62  v  4  ac  -  62 


1          .      2  ax  +  6-V  62  -  4  ac   .   n     ..         .  „. 
log +  C    (4  ac  <  62) 


v  62  -  4  ac        2  ax  +  6  +  V  62  -  4  ac 

11       f     2x  +  7  W2x  +  4)dx  f  __^_ 

Jx2  +  4x  +  5  J  x2  +  4x  +  5  +     J  (x  +  2)2  +  l 

=  log  (x2  +  4x  +  5)  +  3  tan"1  (x  +  2)  +  C. 

J  x2  +  2x  +  1  ~  J  (x  +  l)2^      J  (x  +  l)2 

=  \og(x  +  l)+j±-[  +  C. 

* «      C  dx  If  &  dx  1    .   _  bx   .   n 

13.     I  —  =  j-   I  =  r  sin  L  —  +  (7, 

•J  VoV  -  62x2      b  J  V(ac)2  -  (bx)2       °  ac 


ac 


14.  f    .       X  =  J  log  (&x  +  Vb2x2  +  a2c2)  +  C,  or  sinh"1  -  +  C". 
J  Vb2x2  +  a2c2      b  ac 

15.  f    .     dx  =  I  log  (bx  +  vW  _  a2c2)  +  c,  or  cosh"1  —  +  C 
J  Vb2x2-a2c2      °  ac 

16.  C~7M=  =  -L  f-7-^—  =  4=  sin-  x  \/l  +  C. 
^  v  a  -  6x2       v  6  ^  v  a/6  -  x2       V6  y  a 

17.  f—^=  =  -1=  sin-xV/f  +  C.      18.     f~J^=- 
J  Vs  -  2  x2      V2  V  3  ^  V3  -  4  x2 

19.  f  ,     ^         =  f  #      <**  =  sin-  2-±p-  +  C. 

on       T  dx  •   _,  x  -  1    .    „ 

20.  I  — —  =  sin  l  — ; h  C. 

^  V2  +  2  x  -  x2 

2i.  c      d*     _ .  i  r 

^  v  ax2  +  6x  +  c       v  a  ^ 


V3 

2adx 


V(2ax  +  6)2  +  4ac  -  62 
=  -^  log  (2  ax  +  6  +  2  Va  Vax2  +  6x  +  c)  +  C 

f    .     dx  =  log  (x  +  a  +  Vx2  +  2ax)  +  C. 

^  vf  -}-  2 ax 


,  EXERCISE  XXIII  193 

23.     f    ;  dx        =  -^log  (x  Vo~  +  Vax2  -  b)  +  C. 
J  Vax2  —  b      v  a 

dx  1     /*  2adx 


24.  r      d*      -  =  —  f      — 

J  V  -  ax2  +  bx  +  c       Va  ^  V4  ac  +  62 


(2  ax  -  6)2 


1      .     ,     2ax  -b      ,    _ 
— r-  sin  x  r  +  C. 

Yd  V4ac  +  62 


25.  p^dx  =  f4±^^=r-i^+  ({i-x^xdx. 

J  Vl-x  J  vl  -  X2  J  Vi  -  x2      J 

=  sin"1  x  -  Vl  -  x2  +  C. 

26.  p£±»  fc  =  rfe  +  S*  =  VJ5TT+  log  (,  +  V5^D  +  C. 

^  v  x  -  1  ■*    v  x2  -  1 

f  dx  r  cdx  1  ,  ex   .    ~ 

27.  I  .  I .  ==  -rsec-x-7-  +  C. 

J  x  Vc2x2  -  a2b2      J  ex  V(cx)2  -  (a&)2      a»  a° 

28.  C       5dx        . 

J  x  V3x2  -5 

29.  f  V*±JJldx  =  f     X.+g      dx  =  sec-1  -+  log  (x+ Vx^O+C. 
J  xVx— a         J  xVx2  —  o?  a 

so.  f2*EEI*.  r  ";z^Ld^  r^^-f-^^ 

J         x  J  x  Vx2  —  a2  ^  vx2  —  a2     ^  x  vx2  -  a2 

=  Vx2  -  a2  -  a  sec"1  -  +  C. 
a 

81.     f— *-,-  r-^*L5-f(^-1)-l^.&- _*+c. 

J  (1  _  X2)§       J  (X"2  _  i)s      J  Vl-x2 

««      f  dx  If  6  dx  1  6x       ~ 

32.  j  ——==  =  T  (  =  T  vers  * h  C. 

J  V2  a&x  -  62x2      h  J  V2  a  (6x)  -  (6x)2      °  a 

oo      C         —dx  1  _,  bx   ,    „ 

33.  I  =  =»  r  covers  x  -r  +  C. 
J  V8  6x-62x2      &  4 

oa      f       dx  _.2x  .  n 

34.  I  _  =  vers  L \-C. 

J  Vax  —  x2  a 

„_      /*  — xdx        _   fa  —  2x  —  a  ,   _   r  (a  —  2  x)  dx     a  /*       dx 

Vax  -  x2 


/•  — xdx        _  fa  —  2x  —  a  ,   _   /*  (a  —  2  x)  dx     a  r 
J  Vax  -  x2     J  2  Vax  -  x2  J  2  Vax  -  x2      2  ^ 


, a         _,  2  x   .   n 

Vax  -  x2  -  5  vers  * h  C. 


36.     f -F^=  -(•—**  -  sin-!  (2-^)  +  C 

J  Vax  -  x2      J  Va2/4  -  (x  -  a/2)2  V      a      / 


z2/4  -  (x  -  a/2)2 


194 


INTEGRAL  CALCULUS 


Note.  —  It  may  be  noticed  that  the  result  for  Example  34  differs 
from  the  last  result  for  Example  36.     The  difference  is  accounted  for 

by  the  values  of  the  constants  of  inte- 
gration.    As  may  be  seen  in  the  figure, 


(¥-) 


.     7T  2X 

+  ~  =  vers  l — , 

_  CL 


that  is, 


2X 

a 

a=  2 '■-.radius  =  1 


arc  BP  +  £ 


arc  OBP. 


Each  result  may  be  verified  by  dif- 
ferentiation according  to  (i)  Art.  116. 
These  results  illustrate  the  statement  at  the  end  of  Art.  117. 

122.  Reduction  Formulas.  —  A  formula  by  which  a 
differential  expression  not  directly  integrable  can  be  re- 
duced to  a  standard  form  or  a  form  easier  to  integrate  than 
the  original  function,  is  called  a  reduction  formula. 

A  general  reduction  formula  that  has  a  wide  application 
and  is  most  useful  in  the  reduction  of  an  integral  to  a  known 
form  is  the  formula  for  integration  by  parts. 

Many  special  formulas  of  reduction  are  obtained  by  apply- 
ing this  general  formula  to  particular  forms. 

123.  Integration      by      Parts.  —  Differentiation      gives 

d  (uv)  -  udv  -{-vdu. 
Integrating, 

=    I  d  (uv)  =    I  udv  +   /  vdu; 

transposing, 

I  udv  =  uv  —    I  vdu.  (1) 

The  formula  (1)  may  be  used  for  integrating  udv  when 
the  integral  of  v  du  can  be  found.  This  method  of  integra- 
tion by  parts  may  be  adopted  when  /  (x)  dx  is  not  directly 
integrable  but  can  be  resolved  into  two  factors;  one,  as  dv, 
directly  integrable;  and  the  other,  as  v  du,  a  standard  form 
or  a  form  less  difficult  than  u  dv. 


uv 


INTEGRATION  BY  PARTS  195 

No  rule  can  be  given  for  choosing  the  factors  u  and  dv 
other  than  the  general  direction  that  the  factor  of  /  (x)  dx 
taken  as  dv  is  first  chosen  as  that  part  directly  integrable, 
and  then  what  remains  whether  one  or  more  factors  must  be 
taken  as  u. 

When  the  function  given  to  be  integrated  contains  more 
than  one  factor  that  is  directly  integrable,  there  is  some 
choice  to  be  exercised  in  selecting  the  factor  dv;  and  in  some 
cases  a  different  choice  may  be  necessary,  if  the  first  choice 
results  in  v  du  being  non-integrable.     It  may  be  that  one  or 

more  applications  of  the  formula  to   I  v  du  will  be  effective. 

The  use  of  the  formula  is  illustrated  in  the  following 
examples,  the  formula  being  written, 

dv 


I  f(x)dx  =   j  u 

=  uv  —   J  vdu.  (1) 

sin-1  -  dx  =  x  sin-1  -  +  v  d2  —  x2  +  C. 
a  a 

x 
Let  dv  =  dx:     then    w  =  sin-1-> 

a 

7  dx 

v  =  x,  du  =     . 

Va2  -  x2 

Substituting  in  (1) : 

sm_i  -  dx  =  x  sin-1 I      , 

a  a      J  Va2  -  x2 

=  zsin-1  -  +  Va2  -  x2  +  C. 
(Compare  example  in  Art.  118.) 

Example  2.  —  /  x  •  cos  x  dx  =  x  •  sin  x  —   f  sin  x  dx 

=  x  sin  x  +  cos  x  +  C. 


196  INTEGRAL  CALCULUS 

Example  3.  — 
/  x2  sin  x  dx  =  2  x  sin  x  —  (x2  —  2)  cos  x  +  C 

/  x2  •  sin  x  dx  =  x2  (—  cos  a;)  +  2  J  cos  x  •  x  dx 

=  x2  ( —  cos  x)  +  2  (x  sin  x+cos  x)  +  C,  by  Ex.  2, 
=  2  #  sin  x  —  (x2  —  2)  cos  x  +  C. 
Example  4.  — 

JV  log  x  <b  =  ^  (log  x  -  -L_)  +  C. 

/xn+1    .           r  xn+1    ^ 
logX'Xndx  = — r-lOgX—     /    — -• 
&                   n  +  1       &         J  n  +  1     x 

-vn+1                         /»     rrn+l 
«=  f— r  log  X  -     /   7 —j^  +  C 

n  +  1  J  (n  +  l)2 


l(^i-Jfil)  +  C- 


Example  5.  —  /  xex  dx  =  xex  —  ex  +  C. 

j  ex'xdx  =  ex*ix2  —  i  f  x2-exdx. 

The  last  form  is  not  so  simple  as  the  original,  indicating 
that  a  different  choice  of  factors  should  be  made.  Another 
choice  gives 

I  x '  ex  dx  =  x  -  ex  —    j  ex-dx. 

=  xex  —  ex  +  C. 
Example  6.  — 

fVa^tfdx  =  IxVtf-x*  +  f  sin-1^  +  C.  (1) 

f  Va1^?  •  dx  =  Va^^x1-  x  +   f-^= 
J  J  Va2  —  x2 

=  xVa^x~2+  fa2-^Lx2)dx 
J       Va2  -  x2 

=  x  Va2  -  x2  +  a2  sin"1  -  +C"  -   fVtf^tfdx. 
Transposing  the  last  term  and  dividing  by  2  gives  (1). 


INTEGRATION   BY  PARTS  197 

Example  7*  — 
f  V^+^dz  =  ^xVx2+a2+^a2\og(x+VxT+a2)+C     (2) 

=  \x  VxJ+a2+  I  a2  sinh-1  -  +C".  (2') 

_  Z  a 

Example  8.*  — 
fVx*-d*dx=  lxVx>-a2-  j=aHog(x+Vx>-a?)+e    (3) 

=  \x  Vx^tf-  I  a2  cosh"1  -  +C.  (3') 

—  Z  a 

Note.  —  The  integrals  of  the  three  last  examples  are  of 
sufficient  importance  to  be  considered  as  standard  forms. 

Exam-pie  9.  — 
f  x  I — - —  jdx  =    j  x  •  sinh  xdx  =  x  cosh  x  —  sinh  x  +  C. 

/  x  •  sinh  xdx  =  X-  cosh  x  —    I  cosh  x>dx 
=  x  cosh  z  —  sinh  x  +  C. 

EXERCISE   XXIV. 

Verify  the  following  by  j  udv  =  uv  —  j  vdu. 

1.  f  cos"1  -  dx  =  x  cos"1  -  -  Va2  -  x2  +  C. 
J  a  a 

2.  f  tan"1  -  dx  =  x  tan"1  -  -  \  a  log  (a2  +  x2)  +  C. 

3.  f  cot"1  -dx  =  x  cot"1  -  +  I  a  log  (a2  +  x2)  +  C. 

4.  j  log  x  dx  =  x  (log  x  —  1)  +  C. 

5.  f  x  log  x  dx  =  \  x2  (log  x-h)  +C. 

6.  f  x3  log  x  dx  =  I  x4  (log  x  -  I)  +  C. 

*  By  same  method  as  in  Example  6. 


198  INTEGRAL  CALCULUS 

7.  f  x  (<?*  +  e~ax)  dx  =  -  (ecx  -  e~ax)  -  i  (e«  +  e"^)  +  C. 

8.  f  ez  sin  xdx  =  \ex  (sin  x  —  cos  x)  +  C. 

Take  w  =  ex  and  apply  the  formula,  then  take  u  =  sin  x,  apply 
formula;  add  results. 

9.  f  x2  cosxdx  =  2x  cos x  +  (x2  -  2)  sin x  +  C. 

10.  f  (log  x)2  dx  =  x  [(log  x)2  -  2  log  x  +  2]  +  C. 

11.  /xMlogx)2dx  =  ^  [(logx)2  _-^Ilogx+  ^-2]  +C. 

12.  §T~r& tan_1  *  =  (x  ~  ^ tan_1  *) tan_1  *~  loS  (Vl  +  x2)  +  C. 

124.   Reduction  Formulas  for  Binomials.  —  By  applying 

the  formula  for  parts  to   /  xm  (a  +  bxn)p  dx  that  integral  may 

be  made  to  depend  upon  a  similar  integral,  with  either  m  or 
p  numerically  diminished. 

There  are  four  such  formulas,  which  are  useful  for  refer- 
ence, but  there  is  no  need  that  they  should  be  memorized. 


/  xm(a 


+  bxn)pdx  = 


xm-n+l  (a  _|_  fon)p+l 


b  (np  +  m  +  1) 
or 


b  {np  +  m  +  1 ) 

a  (m  —  n  +  1)    T         ,     .  7     x    T        ,  4  s 
/  xm~n  (a  +  bxn)pdx,     (A) 


\         ,   V    H ,    ^  ,   1    /  xm  (a  +  bxn)p~l  dx,     (B) 

m?  +  ra  +  1         np  +  m  +  1  J 


or- 


a(m+l)  a(m+l)        J 

or 

—, — tty \~     — 7 — rTV-   /  xm  (a+bxn)p+1  dx.     (D) 

aw(p+l)  aw(p+l)     J 

Formulas  (A)  and  (B)  are  used  when  the  exponent  to  be 


REDUCTION   FORMULAS  FOR   BINOMIALS  199 

reduced,  m  or  p,  is  positive;  (A)  changing  m  into  m  —  n, 
and  (B)  changing  p  into  p  —  1. 

Formulas  (C)  and  (D)  are  used  when  the  exponent  to  be 
reduced,  m  or  p,  is  negative;  (C)  changing  m  into  m  +  n, 
and  (D)  changing  p  into  p  +  1. 

When  any  denominator  becomes  zero  the  formula  is  in- 
applicable, and  the  integral  can  be  obtained  by  some  method 
without  the  use  of  reduction  formulas. 

Formulas  (A)  and  (B)  fail  when  np  +  m  +  1  =0. 
Formula  (C)  fails  when  m  +  1  =  0. 
Formula  (D)  fails  when  p  +  1  =0. 

EXERCISE   XXV. 

1.     f  "f  ^     =  sin"1  -  +  C,  when  w  =  0,  Standard  Form  XVIII. 
J  Va2-x2  a 

C     Xdx     =  C(a*—x*r*xdx=  -Va2-x2-\-C,  whenwi  =  l.    (1) 


Va2-x2 


J  Va2-x2     J 


m 


C    x~_^_  =    fj*  (a2  -  x2)~Ux  =  -  %  Vtf 


VC2-!2 


+  %  sin"1  -  +  C,  when  m  =  2.  (2) 

f   f4*     =    Cx*  (a2  -  x2)~*  ax  =  -  %  Vtf~T* 

J  Va2—x2      J  6 

+  x  a2  J  : ,  when  m  =  3, 

o     J  Va2  —  x2 

=  _|va*-rf-|a*Vtf_rf  +  C,  by  (1).     (3) 

f    f*^     =    f  x4  (a2  -  x2)~* dx=  -^  Vtf=7* 
V  va2— x2      ^  * 

3    „  r     x2e7x 


+  -a2  I     ;  ,  when  m  =  4, 

4      J  Va2  -  x2 

(x5  ,  3a2i\    /- :  .   3  a4    .  _,  x  ,  ^  ,      ,ON 


(4) 


200  INTEGRAL  CALCULUS 


2.  I  xm  Va2  —  x2  dx  =  r-x r-=  I     .  ,  by  (B). 

J  m  +  2  ra  +  2  J  Va2  -  x2 

f  V^T^dx  =  ?  V^T^  4. 12  f   /**       ,  when  m  =0, 
J  2,  I  J  Va2  —  x2 

x  A/-5 ;  .   a2   .  _t05   .    ~    /Compare  Ex. \ 

=  _V^-^+2Sml5  +  C-   (  6,  Art.  123.) 

/,    /-= 7  j        Xs  Va2  -  x2  ,  a2  r     x2dx  ,  _ 

X2  Va2  -  x2  dx  = 3 1-  -r  I      ,  ,  when  m  =  2, 

4  4  •J  v  a2  —  x2 

=  (f-|^)^/^r^  +  i^2sin-^  +  C,  by(2)ofEx.l. 

3.  f^!^  =  ^  VxM^"2  -  f  log(s  +  Vtf  +  tf)+C,    by  (A), 
J  Vx2  +  a2      *  * 

"2     X+a        2Smh    a  +  C'     V  7,  Art.  123.  ) 

4.  f   /<**      l|VF^^  +  |>log(x+V^3tf)  +  C>     by  (A), 
J  Vx2  —  a2      &  * 

.  |  V^^  +  £  cosh-  5+  C.  (  Co7ar«  Ex- 8>) 
2  2  a  \       Art.  123.      / 

/*  o^r^—, — 5j        x3  Vx2  ±  a2    ,   a2    /•    x2dx,  ,      m, 

5.  I  x2  vx2  ±  a2  dx  =  ^ ±  -r   I     ,  ,     by  (B), 

J  4.4./  Vx2  ±  a2 

=  (f  ±1^)"^^  -^2lo8^+  ^±^)+C,  byExs.  3,4. 

6.  f>  Vx^dx  =  ^  ^fp  ±     *      f-^==,  by  (B). 
J  m  +  2  m  4-  2 »/  Vx2  ±  a2 

7      f         rfa;  Vx2  d=  a2 

J  x2  Vx2  ±  a2       T        a2x       "•" 

JW±  *>-**-  £1^±^ 

-  (-1-2  +  2  +  1  r  0      ±  a2)_j  rfx>    by  (C)< 

T  a1  J 


«/    t2  V/i2  _  ^2  GTX 

i 


x2  Va2  -  x2 

x"1  (a2  -  x2)1 


JV2  (a2  -  x2)_i  dx  =  —  _ 

-  -K-l-2  +  2  +  1)   r^       _      -,  dx>        (C)# 
—a2  J 


REDUCTION  FORMULAS  FOR  BINOMIALS         201 


C         dx  Vx2  —  a2         1         _,  x       ~ 

=  (:C2~q!)3  +  ^  sec"1  -  +  C,  by  Standard  Form  XXI. 
2a2x2  2  a3  a 

in      C         dx  1  1       _,*.,  /, 

10.     | j  = .  r  sec  1  -  •+  C. 

•J  x  (x2  -  a2)5  a2  Vx2  —  a2      a  a 

JV1  (x2  -  a2)"^x  =  -  ^fr2"02)   ' 

-  |  JV1  (x2  -  a2r"dx,     by(D), 
_lsec-i-+C. 


a2  Vx2  _  a2       a3  a 

H      C dx        =  x  +  c 

J  (a2  -  x2)*      a2  Va2  -  x2 

C(n2      *\-*L  x(a2-x2)~* 

J(a2-x2)idx=-    2fll(_4) 

+  "32+a2°(-')+1/a;0(a2-^^     by(D)- 

12.     fV2ax-x2  dx  =  ?-^L  V2ax-x2  +  %  sin"1  ^^  +  C, 
J  A  _  a 

or    — —  v  2  ax  —  x2  +  —  vers  x  -  +  C 

£  a  CL 

fVfZ^dx  .  p(2a  -*)»«&.     CAPP'y  (A)  and  (B)) 
»/  ^  \      in  succession.      / 


Or 


fV2  ax  -  x2dx  =  f^a2  -  (x  -  a)2 


dx 


a-  .     ,x  —  a 


V2  ax  -  x2  +  %  sin"1  — -  +  C,  by  Ex.  2, 


2         '   2  a 


x  —  a 
~~ 2" 


V2ax-x2  +  g-2vers-1g  +  C      f     See  Ex.  36,      \ 
V  2  ax      x  +  2  vers     o  +  C  .     ^Exerdge  XXIIlJ 

13.     f  x™  V2ax-x2  dx  =  f  z*H  V2a-x dx  =  -  xm~l(2ax~x^ 
J  J  m  +  2 

+  (2m  +  ^)a  fx™"1  V2ax-x2dx,     by  (A). 


202  INTEGRAL  CALCULUS 

xm  dx  xTO_1  V2  ax  -  x2 


"•/ 


i,/ 


V2, 


(2m -1)  a  r    xm~1dx 


J 


V2ax  -  x2 


dx 


V2 


ax  —  x6 


xmV2ax-x2  (2m-l)oaJ» 

m  —  1       /*  dx 


+ 


oJ   rw-i 


i,/ 


dx 


(2  m  -  1)  a  J  zm-i  V2ax  -  x2 
x 


Va2  -  x2   .     1    . 

+  — ,  log 


,    by(C). 

+  C. 


x*V  a2  -  x2  2  °>2x2  2  a3        v  a2  -  x2  +  a 

-"1("1r23a2+2  +  1)p-1^-x2)-^x)    by(C), 
+  i  f     /*       .    See  Ex.  18. 


Va2  -  x2  ,     1 
2a2x2 

Vx2  +  a2         1 


/dx  _Vx2_+a2 1_,  x c 

x*Vx2  +  a2  2  °<2x2  2  °3        a  +  V&+  a2 

- ll~1~_lt2+1) S** (x2 +ai)~* dx'  by (C)- 

VX2    +  gl  1  /•  dX 


18. 


/: 


dx 


=  -log 


V  a2  -  x2      «        v  a2  -  x2  +  a 


+  C. 


Here  m  +  1  =  0,  therefore  Formula  (C)  fails. 

Let  a2  —  x2  =  22;      .'.      —xdx  =  zdz,  x2  =  a2  —  z2. 

/dx         _  r     dz  1  ,      a  —  z       ~ 

x  Va2  -  x2  ~~J  22-o2"2o     go+z  + 

1   .      a  -  Va2  -  x2      n 

=  rr-  log ,  +  C 

2  a        a  4.  Va2  -  x2 


-log ; 

a       a  +  v  a2  —  x2 


+  C. 


19 


REDUCTION  FORMULAS  FOR  BINOMIALS         203 

•    f     7^_  =  V    7_*  +C. 

J  x  Vx2  +  a2      a        Vx2  +  a2  +  a 

Again  m  +  1  =  0,  and  Formula  (C)  fails.  • 

Let  a2  +  x2  =  z2;     :.     xdx  =  z  dz,  x2  =  z2  -  a2. 

/<fc         _  /•  _dz__        1  .      z  —  a       ~, 
x  Vj2  +  a2     J  z2  —  a2      2  a        z  +  a 

1   .       VxT+~a2  -  a   .   n 

=  j^-  log     . h  C 

2  a     &  Vx2  +  a2  +  a 

--log     ,       *  : +  C. 

a        Vx2  +  a2  +  a 

Note.  —  The  integrals  of  Examples  18  and  19  may  be  considered  as 
additional  standard  forms. 

on      fVtf-x2,  ,- .     .  x  /By  (B)  and\ 

20-    J-^— ^=Va2-x2  +  aloga+Va^+C.(    ^/lgi     ) 


CHAPTER  II. 


DEFINITE  INTEGRALS.     AREAS. 

125.  Geometric  Meaning  of  /  /  (x)  dx.  —  As  the  repre- 
sentation of  an  integral  by  an  area  between  a  curve  and  an 
axis  is  of  fundamental  significance,  and  as  the  effort  to  find 
an  expression  for  the  area  of  plane  figures  bounded  by 

curved  lines  gave  rise 
to  the  Integral  Calcu- 
lus, such  representa- 
tion will  be  given 
further  treatment 
than  illustrated  in  the 
examples  of  Art.  115. 
Let  P2OPi  be  the 
locus  of  y  =  f  (x),  and 
let  the  area  between 
the  curve  and  the 
z-axis  be  conceived  as 
generated  by  the  va- 
riable ordinate  MP  or  y,  as  the  point  (x,  y)  moves  along  the 
curve  and  x  increases.  Let  A  denote  the  area  bounded  by 
the  a>axis,  y  =  /  (x) ,  some  undetermined  fixed  ordinate  as 
MqPq  or  M2P2,  and  the  moving  ordinate  MP. 

Let  Ax  =  dx  be  MM\\  then,  while  A  A  the  actual  incre- 
ment of  the  area  A  is  MPPiMi,  dA  is  MP  DMh  the  incre- 
ment that  A  would  get  if,  at  the  value  M0P0PM,  the  change 
of  A  became  uniform  and  so  continued  while  x  increased 
uniformly  from  the  value  OM  to  OMi.    Hence 

204 


&**]% 


DERIVATIVE  OF  AN  AREA  205 

dA  =  MPDMi  =  ydx=f(x)  dx, 
:.     A  =   J  ydx  =    I  f(x)  dx, 

where  A  'is  indeterminate  so  long  as  the  fixed  ordinate  M0P0 
or  M2P2  is  indeterminate. 

,     dA 


126.   Derivative  of  an  Area.  —  Since  dA 


that  is,  the  derivative  of  the  area  with  respect  to  x  is  the  ordinate 
of  the  bounding  curve. 

This  important  result  may  be  obtained  by  the  method  of 
limits  also,  taking  the  increments  infinitesimal.     Thus, 

A  A  =  MPP1M1, 

AA  >  y  Ax,    and    AA  <  (y  +  Ay)  Ax, 
.*.     y  Ax  <  AA  <  (y  +  Ay)  Ax, 
AA 

dA      v     A  A 

.'.     -j-  =  lim  -r —  =  y, 
dx      Ax=o  Ax       *' 


since 


lim  (y  +  Ay) 

Ax=0 


y,     Ay  =  0    as    Ax  =  0. 


In  case  y  decreases  as  x  increases, 
the  curve  falls  from  P  to  Pi,  and  the 
inequality  signs  are  reversed,  but  the 
result  is  the  same. 

Let  A  be  the  area  between  the  t/-axis 
and  the  curve;  then, 

dA 
dy 


lim  -r — 

Ay=o  Ay 


=  x,     or     dA  —  x  dy, 
=   I  xdy. 


Here,  the  derivative  of  the  area  with  respect  to  y  is  the  abscissa 
of  the  bounding  curve. 


206  INTEGRAL  CALCULUS 

127.  The  Area  under  a  Curve.  —  Let  the  curve  y  =  f  (x) 
of  Art.  125  bey  =  x2. 

Let  OM0  =  a  and  OMi  =  b;  then 

MoPoPM  =  A=    fx2dx  =  ^  +  C. 
As  the  area  is  measured  from  x  =  a, 

...    4  =  o  =  !  +  c,   c=-|, 

£3        tt3 

•'•     A^=3-3'  (1) 

where  Ax  is  the  variable  area  M0P0  J°M.     Making  x  *=  6  in 
(1)  gives 

MoPoPiM,  -  A6  -  |  -r  J-  (2) 

The  usual  notation  is 

,         /"6  2  ,        z31&      63      a3  ,oA 

128.  Definite  Integral.  —  In  general,  when  b  >  a  the 
increment  produced  in  the  indefinite  integral  F  (x)  -f-  C  by 
the  increase  of  x  from  a  to  6  is 

F(b)+C-  (F(a)  +  C)  =F(6)  -F(a). 

This  increment  of  the  indefinite  integral  of  /(#)  dx  is  called 
"the  definite  integral  of  f(x)  dx  between  the  limits  a  and  6," 

and  is  denoted  by  f  f  (x)  dx.    Hence, 


s. 


bf(x)dx~F(b)  -F(a), 

a 


The  operation  is  that  of  finding  the  increment  of  the  indefinite 
integral  of/  (x)  dx  from  x  =  a  to  x  =  b,  where  b  is  called  the 
upper  or  superior  limit,  and  a  the  lower  or  inferior  limit, 
although  they  are  more  precisely  termed  "end  values"    of 


DEFINITE  INTEGRAL  207 

the  variable,  as  they  are  not  "limits"  in  the  usual  sense  of 
the  word.     If  the  upper  end  value  is  variable,  then 

fXf(x)  dx  =  F  (z)T  =  F(x)-  F(a). 

When  the  lower  end  value  a  is  arbitrary,  —  F  (a)  may  be 
represented  by  an  arbitrary  constant  C,  hence 


£ 


f(x)dx  =  F(x)+C. 

*J  a 

Since 

r'f(xydx  =  F(x)+C, 


/■ 


an  indefinite  integral  is  an  integral  whose  upper  end  value 
is  the  variable  and  whose  lower  end  value  is  arbitrary. 

Hence,  when  the  integral  is  represented  by  an  area  and 
the  area  is  known  for  some  value  a  of  x, 

A=  ff(x)dx  =  F(x)-F(a), 

where  C  is  —F  (a),  and  the  area  A  is  determinate. 

If  the  area  under  the  curve  y  =  x-  (Art.  127)  be  reckoned 
from  z  =  0;   when  x  is  zero,  A  is  zero,  therefore,  C  is  zero 

and  Ax  =  j ,  the  area  of  OPM.  (3) 

(See  Example  4,  Art.  115.) 

Av  =  fy* dy  =  lyi  =  l x\  the  area  of  OPX.        (3') 

Making  x  =  a  in  (3)  gives 

a3 
A  a  =  -w ,  the  area  of  OP0M0)  (4) 

o 

e 

and  making  x  =  b,  gives 

Ab  =  ^,  the  area  of  OPJIl  (5) 

o 


208 


INTEGRAL  CALCULUS 


x3~|a      a3  x3~]h 

H-3'  "  si 


'I 


¥  -  a2, 


As  definite  integrals, 
A  =    I    x2dx 

If,  in  Example  3  of  Art.  115,  the  area  A  is  from  x  =  a  to 
x  =  6;  then 

A  =    /    2xdx 

the  area  of  a  trapezoid.  It  may  be  noted  that,  when  the 
integral  is  " between  limits,"  it  is  not  customary  to  write 
the  constant,  as  it  will  be  eliminated. 

129.  Positive  or  Negative  Areas.  —  The  area  under  or 
above  a  curve  y  =  f  (x),  from  x  =  a  to  x  =  b,  will  be  posi- 
tive or  negative  according  as  y  is  positive  or  negative  from 
x  =  a  to  x  =  b;  hence,  when  the  curve  crosses  the  x-axis, 
the  areas  are  gotten  separately,  otherwise  the  result  will  be 
the  algebraic  sum  of  the  areas  and  may  be  zero,  since  the 
areas  above  and  below  the  axis  may  for  some  curves  be  equal. 
For  example,  the  areas  for  the  curve  of  sines  or  of  cosines 
as  shown  in  Art.  140  illustrate  the  principle. 

130.  Finite  or  Infinite  Areas  —  '  'Limits  "  Infinite. — 
From  the  geometrical  meaning  of  an  integral  it  follows  that 
/  (x)  dx  has  an  integral  whenever  /  (x)  is  continuous,  hence 

the  end  values  a  and  b  are  in 
general  taken  so  that/  (x)  will  be 
finite,  continuous,  and  have  the 
same  sign,  from  x  =  a  to  x  =  b. 
If  x  =  b  =  oo ,  then 


Jrv>  r*x=b 

f(x)dx  =  lim   I       }{x)dx, 
a  6=oo  *Jx  =  a 


-  where  the  limit  may,  or  may  not, 
exist.  When  the  limit  of  the  in- 
tegral is  finite  the  total  area  is  found;  but  if  as  b  becomes 
infinite  the  integral  becomes  infinite,  then  no  limit  exists  and 
the  area  up  to  x  =  b  becomes  infinite  as  b  becomes  infinite. 


FINITE  OR  INFINITE  AREAS  —  LIMITS  INFINITE      209 

Example  1.  — 

A  =    /     —  dx  =  lim  /    --  cfo  =  lim =  1. 

J l      X1  b=*>  Jl    X£  &=«  L        ZJl 

Limit  exists  although  as  x  approaches  zero,  f(x)  =  —2  be- 

comes  infinite. 

A  =    I     -=  dx  =  hm   /    —  da;  =  lim > 

Jo     x2  b=*Jo  x2  b=«  L      ^Jo 

which  does  not  exist,  since =  ao  ;  that  is,  the  area 

up  to  x  =  b  becomes  infinite  as  b  becomes  infinite. 

A  =    I    -=  dx  =  lim   /    -rrfx  =  lim > 

Jo    X2  a=oJa    X2  a=0  L        ^Ja 

which  does  not  exist,  since =  go  ,  the  area  becoming 

infinite  as  x  approaches  zero. 


-2.0         -1.5  -10  -0.5 


Example  2.  — 

A  =    /  X  e* dx  =  Hm     I    ^dx  =  lim    ex      =  ea 

J  — x  o=  —  x  t/a  a  =  — ^L      Ja 

is  total  area  under  curve  up  to  ordinate  at  P  (x,  y). 


210  INTEGRAL  CALCULUS 

Area  to  right  of  y-axis 

=  OMPB  =  A=    f  exdx  =  e*T  =  ex 
Jo  Jo 

Area  to  left  of  y-axis 

=    /      exdx  =  lim     /    ex  dx  =  lim     ex 

%J  —  co  a=  —  oo  *J  a  a  =  —  oo  |_     J 


0 

=   1. 


Note.  —  When  y  =  /  (x)  =  e*,  ^/,  the  function,  is  the  ordi- 
nate, equals  the  slope  at  the  end  of  the  ordinate,  and  may 
represent  the  total  area  under  the  curve  up  to  the  ordinate. 
(See  Art.  138.) 

131.   Interchange  of  Limits.  —  Since  the  definite  integral 


£ 


bf(x)dx  =  F{b)-F(a); 


it  follows  that 

I  f(x)dx=  -    I   f  0)  dx, 

Ja  Jb 

since  the  second  member  is  —  [F  (a)—  F  (&)]=  F(b)  —  F(a). 
It  follows  also  that  the  definite  integral  is  a  function  of  its 
limits,  not  of  its  variable;  thus 

Jf  (y)  dy  has  the  same  value  as  /   /  (x)  dx, 
a  J  a 

each  being  F(b)  -  F(a). 

The  algebraic  sign  of  a  definite  integral  is  changed  by  an 
interchange  of  the  limits  of  integration,  and  conversely. 

132.  Separation  into  Parts.  —  A  definite  integral  may  be 
separated  into  parts  with  other  limits  or  end  values.     Thus, 

fbf(x)dx=    fCf(x)dx+   fbf(x)dx.  (1) 

Let  the  curve  y  =  /  (x)  be  drawn  and  the  ordinates  APX, 
BPh  CP2,  be  erected  at  the  points  for  which  x  =  a,  x  =  b, 
X  =  c. 


MEAN  VALUE  OF  A  FUNCTION 


211 


Since  area  APP^B  «=  area  APP,C  +  area  CP2P1B,    (1) 
follows. 


A  C      B      C 

If  OC  =  c',  and  CrP$  is  the  corresponding  ordinate,  then, 
area  APP^B  =  area  APPZC  -  area  £PiP3C"; 
and  hence 

Pf(x)dx  =    fCf(x)dx-   fCf(x)dx 

=  ff  (x)  dx  +  flJ(x)  dx,  by  Art.  131. 

Note.  —  It  may  be  seen  that 

Jf(x)dx=    I    f  (a  —  x)  dx, 
0  t/0 

for  each  is  F(a)  -  F(o).    Thus, 

-    Pf(a  -x)d{a-x)=  -F(a-  x)T  =  F(a)-F  (o) 

=    Pf(x)dx. 

Jo 

133.   Mean  Value  of  a  Function.  —  The  mean  value  of 
/  (x)  between  the  values  /  (a)  and  /  (b)  is 


./>> 


dx 


Let  area  APPiB  represent  the  definite  integral   /   /  (x)  dx. 


212 

Then 


INTEGRAL  CALCULUS 


fbf(x)dx  =  areaAPPi# 

=  area  of  a  rectangle  with  base  A  B  and  height 

greater  than  AP  but  less  than  BPi 
=  AB  •  CP9 
=  (b  —  a)f(c),  where  OC  =  c. 


Hence 


f(c)  = 


f>< 


x)  dx 


b  —  a 


where/  (c)  is  the  mean  value  of  f  (x)  for  values  of  x  that  vary 
continuously  from  a  to  b. 


The  mean  value  may  be  denned  to  be  the  height  of  a 
rectangle  which  has  a  base  equal  to  b  —  a  and  an  area  equi- 
valent to  the  value  of  the  integral. 

Example  1.  —  To  find  the  mean  value  of  the  function  Vx 
from  x  =  1  to  x  =  4.      Let  OP  Pi  be  the  locus  of 

y  =  v'z, 


/(c) 


OA  =  1     and    OB  =  4. 

2       »7 

I    x2dx       ix^\ 


-> 


H  = 


14 


4-1  3  9 1~      ~J       9 

'  1£  =  1.56|  =  CP',  mean  value. 
x  =  (V)2  =  W  =  2.42  =  OC. 


EVALUATION  OF  DEFINITE  INTEGRALS  213 

Example  2.  —  To  find  the  mean  value  of  sin  0  as  0  varies 
from  0  to  7r/2,  or  from  0  to  w. 


T 


f    sind  dO 

"Ho         1         2 

tt/2-0 

7T/2                 7r/2          7T 

/  * sinddd 

—  cos  0 

-*°  —       —  0  fi^fifi 

7T-0 

—      —  u.uouu. 

7T                     7T 

0.6366. 


Example  3.  —  To  find  the  mean  length  of  the  ordinates  of 
a  semi-circle  of  radius  a,  the  ordinates  for  equidistant  in- 
tervals on  the  arc. 


J    a  sin  0  dd       —  a  cos  0         n 
o  Jo      2  a 


-0 


0.6366  a. 


Example  4.  —  To  find  the  mean  length  of  the  ordinates  of 
a  semi-circle  of  radius  a,  the  ordinates  for  equidistant  inter- 
vals along  the  diameter. 


a -(-a)  2  2  aJ-0        2       2a 


Va2  -  z2  dx 

=  7}X  Va2  —  x 

"2a 


=  ja  =  0.7854  a. 
4 

134.   Evaluation  of  Definite  Integrals. 


EXERCISE   XXVI. 

.n+i  -|  b        frn+i  _  an+i 


i.  f^ci*-— T-— ^- 

Ja  n  +  ljo  n  -f- 1 

2.  J24x3dx  =  x4]2=  16- 1  =  15. 

3.  j^(f  -  ^)  =  log  x  +  log  (2  -  z)]*  =  log  (2  x  -  a?) J* 


=  log(2x  -x2). 


214  INTEGRAL  CALCULUS 

X*>  1  "l°°         1 

e~axdx  =  -  -e~ax        =  -• 
a         Jo        a 

TT  TT  IT 

5.  C  ^r-ndd  =  C  cos-20sin0d0  =  sec0~|    =  V<2  -  1. 
Jo    cos20  Jo  Jo 

a      ra       dx  .    xla     7T 

6.  I       ,  =  arc  sin-       =  =• 
Jo    Va2  —  x2  a-Jo      2 

„      ra     dx  1  x~\a       ir 

8.     r2^»-Rr. 
Jo     v  2  r  —  y 

r<*>    xdx        1        ,        ."I00      x 

10-  J.  r+^  =  2arctan*l  -r 

11.  f  ,e,   ^L  =  arc  tan  ex      =  arc  tan  ex  —  \- 
Jo   1  +e2X  Jo  4 

-~      r1    .  •--  ,  r^l  -  cos  2  0  .„      it 

12.  J     sin2 Odd  =  J     ■ = rf(?  =  T 

-«      f*      „     „        r*l  +  cos  2  0  ,„      TT 

13.  J    cos2 Odd  =  J     — E- ^ rf0  =  j- 

Note.  —  Considering  the  areas  between  the  axes  and  the  graphs: 

TT  IT 

sinn xdx  =   \     cosn a; cfo,  where  n  is  positive, 

o  Jo 

Jsinn  x  dx  =  2   |    sinn  x  dx,  where  n  is  positive, 
o  Jo 

IT 

J    cos"  .t  dx  =  2    I     cos"  x  dx,  if  n  is  an  even  integer, 
0  ^0 

but  =  0,  if  n  is  an  odd  integer. 

135.   Areas  of  Curves.  —  As  has  been  shown,  the  formu- 
las in  rectangular  coordinates  are 

A  =    I  ydx    and    A  =    j  xdy. 


AREAS  OF   CURVES 


215 


(a)  Let  A  denote  the  area  between  the  curves  y  =  f  (x) 
and  y  =  F(x);  let  x  =  031,  dx  =  PE;  then,  the  variable 
area  A  =  P,P'P  and  dA  =  PP'DE  =  (/  (x)  -  F  (x))  dx; 

:.     area  P.PT.P  =  A=    I *\f  (x)  -  F  (x))  dx, 

where  the  points  of  intersection  are  (x0,  y0)  and  (xi,  y{). 

If  the  locus  of  y  =  F  (x)  is  the  x-axis  and  x0  and  X\  are  a 
and  b,  this  formula  reduces  to 


4  =  J> 


x)  dx. 


(b)    In  polar  coordinates, 
A  =  h 


s 


p-dd. 


For  let  P0  be  any  fixed  point  (p0,  0O)  and  P  (p,  0)  any  variable 
point. 

Consider  the  area  PGOP  to  be  generated  by  the  radius 
vector  p  as  0  increases  from  0O,  and  denote  it  by  A.  With 
OP  as  a  radius  draw  the  arc  PD  and  let  dd  =  Z  POPi ;  then 

dA  =  OPD  =  ip-pdd  =  i  p2  dd, 

the  increment  of  A,  if  uniform,  as  in  a  circle. 

/.     A  =  |  /Vd0,     or    i  =  |   Tp2^,  if  <?o  =  0. 

Je0  «^o 


216  INTEGRAL  CALCULUS 

By  method  of  limits,  increments  infinitesimal: 


p2  A0  <  AA  <  §  (p  +  Ap)2  A0,  where  AA  =  OPPh 

:     -r-  =  lim  -r-r-  =  s  p2,  since  Ap  =  0  as  A0  =  0. 

ad       &e±o  A0       J 


EXERCISE   XXVH. 

1.  Find  the  area  between  the  parabola  y2  =  4  x,  the  x-axis,  and  the 
ordinate  at  any  value  of  x;  from  x  =  1  to  x  =  4. 

2.  Find  the  area  between  the  parabola  x2  =  4  y,  the  ?/-axis,  and  the 
abscissa  at  any  value  of  y\  from  y  =  1  to  y  =  9. 

3.  Find  the  area  between  the  two  curves  y2  =  4  x  and  x2  =  4  y. 

4.  Find  the  area  between  the  cubical  parabola  4  y  =  x3,  and  the 
x-axis  from  x  =  0  to  x  =  2;  from  re  =  Otox  =  —  2;  from  x  =  —2  to 
x  =  2. 

5.  Find  the  area  of  the  semi-cubical  parabola  4  y2  =  z3,  bounded  by 
the  double  ordinate  at  x  =  4. 

6.  Find  the  area  between  the  line  y  =  x,  and  the  curve  4  y2  =  x3, 
in  the  first  quadrant. 

7.  Find  the  area  included  between  the  parabola  x1  +  y*  =  a%  and 

a2 
the  axes  of  X  and  F.  Ans.  y 

6 

8.  Find  the  area  bounded  by  the  curve  ?/  (1  +  x2)  =  x,  and  the  line 
y  =  I  x.  Ans.  log  4  —  £. 

9.  Find  the  area  included  between  the  parabola  x2  =  4  ay,  and  the 
witch  y  (x2  +  4  a2)  =  8  a3.  Ans.  (2  tt  -  $)  a2. 

10.  Find  the  area  bounded  by  the  witch  y  (x2  +  4  a2)  =  8  a3  and 
its  asymptote  y  =  0. 


Area  =  2  I        „   ,    .    9  =  8  a2  arc  tan  —       =4  ira2. 
Jo     x2  +  4a2  2aJo 

11.  Find  the  area  bounded  by  the  hyperbola  xy  =  1,  its  asymptote 
y  =  0,  and  the  lines  x  =  1  and  x  =  n.  Ans.  log  n. 

When  n  =  qo,  log  n  =  oo  ;  hence  the  limit  does  not  exist,  and  the 
area  between  the  hyperbola  and  an  asymptote  is  infinite.  Since  the 
area  is  the  Napierian  logarithm  of  the  superior  limit,  Napierian  loga- 
rithms are  sometimes  called  hyperbolic  logarithms. 


AREAS  OF  CURVES  217 

12.   Find  the  area  of  the  lemniscate  p2  =  a2  cos  2  0. 
4  a2  A 


Area  =  ^  J*  cos20d0  =  a2. 


13.  Find  the  area  of  the  cardioid  p  =  2  a  (1  —  cos  0). 

Area  =  2  a2  f  ^(1  -  cos  d)2dd  =  Qira2. 

Jo 

14.  Find  the  area  of  the  circle  p  =  2  a  sin  6. 

Area  =  2  a2  f  Tsm20d0  =  7ra2. 

Jo 

15.  Find  the  area  of  the  circle  p  =  2  a  cos  0. 

Area  =  2  a2  f  Xcos20d0  =  ira2. 

Jo 

16.  Show  that  the  difference  between  the  areas  of  the  two  circles 
above,  from  0  =  0  to  0  =  tt/4,  is  a2;  also  that  the  area  of  one  circle 
intercepted  by  the  other  is  twice  the  area  of  the  first  circle  from  0  =»  0 
to  0  =  ?r/4. 

17.  Find  the  area  of  the  part  of  the  circle 

p  =  a  sin  0  +  b  cos  0,  from  0  =  0  to  0  =  x/2. 


18.   Find  the  area  of  one  loop  of  the  curve  p  =  a  sin  2  0. 

7T  X 

.1 

'0 


1    r1-  a-   r^  to? 

Area  =  ij     a2sin2  2ddd  =  j  J     (l-cosid)dd.  Ans.  —  • 


19.  Find  the  area  between  the  first  and  second  spire  of  the  spiral  of 
Archimedes  p  =  ad. 

a2    /*4x  a2    /*2x 

Area  =  %  f     02  d»  -  %  I      02tf0  =  8aV. 

I     Jl  W  6     Jo 

20.  Find  the  area  of  one  arch  of  the  cycloid  x  =  a  (0  —  sin  0), 
y  =  a  (1  —  cos0). 

J  1*9  xa  /*2  x 

yds  =  a2  (1  -  C03  6)2dd, 

0  «^0 

(when  x  =  0,  0  =  0;  and  when  x  =  2ira,  0  =  2  r,  and  dx  =  a  (1  — 
cos  0)  dB) 

=  a2   f 2T  (1  -  2  cos  0  +  cos2  0)  d0  =  3  xa2; 
that  is,  the  area  is  three  times  that  of  the  generating  circle. 


218  INTEGRAL  CALCULUS 

21.   Determine  the  area  of  the  circle  x2  +  y2  =  a2. 
Area  =  4:fydx  =  ij    Va2  —  x2  dx. 

Using  the  parametric  equations  of  the  circle,  x  =  a  cos  0,  y  «=  a  sin  0, 
where  0  is  the  variable  parameter,  gives  dx  =  —a  sin  Odd.  Sub- 
stituting the  values  of  y  and  dx  gives: 

Area  =  4  f    Va2  -  x2  dx  =  -4  f  a2sin20d0, 
(when  x  =  a,  0  =  0;    E  =10, 0  =  |1_ 

IT 

a2 sin2 Odd,  by  Art.  131 

o 

2  T-  -  sm  2  g~]  2  /Compare  Ex.  14  above \ 
'         \_2  4     Jo  \  and  Ex.  12,  Art.  134.  / 

=  7ra2. 

To  get  f  Va2  —  x2  dx,  the  indefinite  integral;  let  x  =  a  sin  <f>;  dx  =  a 
cob  <f>d<f>;  then, 

j  Va2  —  x2  dx  =  a2   C  cos2  0  d<£,  where  <f>  is  the  complement  of  0  above, 

=  f /(1  +  COs2</>)d</) 

=  ~  U+  \  sin2</>)  +  C  =  |  (0  +  sin*cos*)+  C 


f^  +  IVT^  +  c, 


the  indefinite  integral.     Compare  Ex.  6,  Art.  123. 

Area  -  4  f  °  Va2-x2  dx  =  4  1"^  sin"1  -+£  Va2-.r2 + C~\ °  =  «*,  as  above. 
./o  L^  «    ^  Jo 


6  7ra2 


Corollary.  —  Area  of  Ellipse  =  4  -  f    Va2  -  x2  dx  ■  4-  ^f- 


TO  FIND  AN  INTEGRAL  FROM  AN  AREA 


219 


136.  To  find  an  Integral  from  an  Area.  —  An  integral 
may  be  found  from  an  area,  when  it  can  be  gotten  geomet- 
rically from  the  figure. 

Example  1 .  —  Find    /  Va2  —  x2  dx  from  the  figure  of  the 

circle  y  =  Va2  —  x2. 

Area  =  BOMP  =  BOP  +  OMP  =  \  a2tf>  +  i  xy 


M      A 

If  the  initial  ordinate  is  not  OB  and  is  undetermined,  then, 
Area 


=  /v* 


x2  dx  =  -  \^a? 


x2  +  — sin-^  +  C, 


as  above,  C  being  independent  of  x  and  indefinite  when  the 
initial  ordinate  is  undetermined. 

Example  2.  —  Find   I  V2ax  —  x2dx,  using  circle 

y  =  V2ax  —  x2. 

Area  =  OBPM  =  OBPC  +  PCM 

a 


a" 


V 


2VerS_1a+      2 


x  —  a    / 


V2ax 


220  INTEGRAL  CALCULUS 


or  Area  =  CBPM  =  BPC  +  PCM 


a2  .     ,  x  —  a      x  —  a   /= 

= —  v  2  ax  —  x2. 


=  2sin 


If  the  initial  ordinate  from  which  area  is  reckoned  is 
undetermined,  then 

fV2ax-x2dx  =  ^=-?  V2ax-x2  +  \  a2  sin"1  ^—^  +  C, 

*7  Z  ad 


ral 


where  C  =  —r- ,  if  Area  =  0,  when  x  =  0; 


or 


x  —  a 


V2ax-x2  +  \  a2  vers_1  *  +  C'> 


where  C"  =  0,  if  Area  =  0,  when  x  =  0. 
As  may  be  seen  in  the  figure, 


sin-1  - — -  +  s  =  OCP  =  vers-1  -  \ 
a         2  a 


that  is, 

a2 I  .     .  x  —  a  ,  7r\      a 
2lSln    ^T~  +  2 J  =  2 


rc  —  a   .  7ra-      a' 


^  sin-1 

2  a 


+ 


vers' 


(See  Note  at  end  of  Exercise  XXIII.) 
Either  result  gives 


lV2ax- 


x2dx  =  lira2  =  areaofOBA. 


AREA  UNDER  EQUILATERAL  HYPERBOLA         221 

Example  3.  —  Find   /  {mx   +   b)   dx,  by  means  of  line 

y  =  mx  +  b. 

Area  =  OMPB  =  BDP  +  OJ/DB 

=  \  x  •  ??l£  +  #  •  6 

rar2   .  , 
=  —zr-  +  bx. 


If  the  initial  ordinate  is  not  OB,  and  is  undetermined,  then 


/ 


mx3 


(mx  +  b)dx  =  '^r- 


137.   Area  under   Equilateral  Hyperbola.  —  As  in  the 

figure  of  the  circle  y  =  Va2  —  x2, 


Xx  1  1 

Va2  —  x2  dx  —  jz  x  Va2  —  x2  +  ■=  a2 


sin-1  - 


expresses  the  area  BOMP  and  a2  sin-1  -  is  represented  by 
twice  the  area  of  the  circular  sector  BOP ;  so 


1 


1 


PVtf  +  tfdx  =^x  Va^+tf  +  ^ a2 sinh-1  -  (Ex.  7,  Art.  123) 
may  be  shown  to  express  the  area  AOMP  under  the  equi- 
lateral  hyperbola  y  =  v  a2  +  x2,  and  a2  sinh-1  -  to  be  repre- 
sented by  twice  the  area  of  the  hyperbolic  sector  A  OP. 


222  INTEGRAL  CALCULUS 

To  get 

jVa2-\-x2dx;    let    x  =  asinh<£,  dx  =  acosh<t>d<f>; 

then, 

/  *Va2  +  x2  dx  =  a2  I  cosh2  <f>  d<f>  »  ^  (</>  +  sinh  <t>  cosh  0) 


=  ^VaHs2+-0fl2  sinh"1  -: 
2d  £i  o> 


as  also  in  Ex.  7,  Art.  123. 


If  x  =  a  cosh  </>  and  dx  =  a  sinh  </>  d<£  be  substituted  in 

l  Vx2  -  a2dx; 
len, 


£ 


Vx2  —  a2  dx  =  ~x  Vx2  —  a2  —  7T  coslr 


1  x 

(as  in  Ex.  8,  Art.  123),  and  ^a2  cosh-11  will  be  represented 

by  the  area  of  a  sector  of  the  equilateral  hyperbola 

y  =  Vx2  -  a2. 


SIGNIFICANCE  OF  AREA  AS  AN  INTEGRAL       223 

138.  Significance  of  Area  as  an  Integral.  —  The  units  of 
the  number  represented  by  A  as  the  measure  of  an  area  will 
depend  upon  the  units  chosen  for  the  abscissa  and  the 
ordinate.  If  the  unit  for  x  be  one  inch  and  that  for  y  be  ten 
inches,  then  a  unit  of  A  would  represent  ten  square  inches; 
if  on  the  graph  the  unit  of  x  is  one-tenth  of  an  inch  and  the 
unit  of  y  is  one  inch,  these  representing  one  and  ten  inches 
respectively,  an  area  on  the  graph  of  one-tenth  of  a  square 
inch  will  represent  the  area  of  ten  square  inches. 

The  integrals  represented  by  areas  may  be  functions  of 
various  kinds,  such  as  lengths,  surfaces,  volumes,  velocities, 
accelerations,  weights,  forces,  work,  etc.  Hence,  the  physi- 
cal interpretation  of  the  area  will  depend  upon  the  nature 
of  the  quantities  represented  by  abscissa  and  ordinate. 

(a)  If,  in  the  figure  of  Art.  125,  the  ordinate  represents 
velocity  and  the  abscissa  represents  time,  then  the  area 
represents  distance;  and,  since  velocity 

ds 

where  a  is  acceleration, 

dA 

-—=  v  =  at. 
dt 

Hence,         A  =  f  atdt  =  hat2  +  C  =  M0P0PM, 

and  since  s  =  \  at2  -+•  s0)  C  is  s0,  initial  distance  or  area;  and 
the  number  of  square  units  of  A  ( =  M0P0PM)  will  equal  the 
number  of  linear  units  of  distance  passed  over  by  a  moving 
point  in  the  time  t  =  M0M. 

(b)  If  the  ordinate  represents  acceleration  and  the  ab- 
scissa still  represents  time,  then  the  area  represents  velocity; 
and  since  acceleration 

_  dv     dA  _     _  dv 
a~dt'     dt~a~dt' 


224  INTEGRAL  CALCULUS 

Hence, 

A  =    Cdv  =  fadt  =  at  +  C  =  MoPoPM, 

where  a  is  constant  acceleration,  and  since  v  =  at  +  v0,  C  is 
Vq,  initial  velocity  or  area;  and  the  number  of  square  units 
of  A  will  equal  the  number  of  units  of  velocity  acquired  by  a 
moving  point  in  the  time  t  =  M0M. 

(c)  If  the  ordinate  represents  a  force  acting  in  a  constant 
direction,  and  if  the  abscissa  represents  the  distance  through 
which  the  force  has  acted,  then  the  area  A  =  MoPoPM 
will  represent  the  work  done  by  the  force  acting  through  the 
distance  represented  by  M0M.  If  the  force  is  constant  in 
magnitude  as  well  as  in  direction,  the  area  will  be  a  rectangle, 

dA  r 

since  -=-  =  F,  constant,  gives  A  =    l  Fds  =  Fs  +  C. 

Whether    the    force    is    constant    or    variable    the    area 

A  =    I  F  ds  represents  the  work  done,  the  area  being  that 

under  the  graph  of  the  equation  y  =  f  (F),  representing  the 
force.  If  the  force  is  not  constant  in  direction,  the  area  will 
still  represent  the  work,  provided  the  ordinate  represents 
the  component  of  the  force  along  the  tangent  to  the  path  of 
its  point  of  application. 

By  means  of  certain  contrivances  the  curve  y  =  f  (F)  may 
be  plotted  mechanically  by  the  force  itself,  as,  for  example, 
in  the  steam  engine  by  means  of  the  indicator.  Having  the 
curve,  the  mean   force  may  be  easily   found;   it   is  given 

Fds     MqM,  the  area  divided  by  the  distance  through 

which  the  force  acts. 

The  area  may  be  read  off  at  once  by  the  polar  planimeter, 
and  the  work  done  found  directly. 

It  is  manifest  that  a  function  may  be  represented  graphi- 
cally either  by  the  ordinate  of  a  curve  or  by  the  area  under  a 
curve;  if  the  ordinate  is  made  to  represent  the  function,  the 


by/ 


AREAS  UNDER  DERIVED  CURVES 


225 


slope  of  the  curve  is  the  derivative  of  the  function;  if  the 
area  under  a  curve  is  taken  to  represent  the  function,  then 
the  derivative  of  the  function  is  the  ordinate  of  the  curve, 
since  the  ordinate  is  the  derivative  of  the  area. 

It  is  usually  preferable  to  represent  by  the  ordinate  *  that 
which  in  the  investigation  is  mainly  under  examination; 
therefore,  if  this  is  the  derivative,  the  latter  method,  where 
the  area  is  the  function  and  the  ordinate  is  the  derivative, 
should  be  used  rather  than  the  former  method,  which  is  to 
be  used  when  the  function  is  mainly  under  consideration. 
The  function  e*  is  exceptional,  in  that  the  ordinate  repre- 
sents the  function,  the  slope  of  the  curve,  and  the  area  under 
the  curve.     (See  Example  2,  Art.  130.) 

139.  Areas  under  Derived  Curves.  —  It  has  been  shown 
(Art.  84,  figures)  by  drawing  the  graphs  of  a  function  and 
its  successive  derivatives  that  the  variation  of  the  function 
is  exhibited  to  advantage. 


It  may  now  be  seen  that  the  area  under  any  derived  curve 
is  represented  by  the  ordinate  of  its  primitive  curve.  Thus 
the  area  under  the  graph  of  y  =  f  (x)  is  represented  by  the 
ordinate  of  y  =  /  (x),  that  under  the  graph  of  y  =  J"  (x)  by 

*  Irving  Fisher,  A  Brief  Introduction  to  the  Infinitesimal  Calculus. 


226  INTEGRAL  CALCULUS 

the  ordinate  oiy  =  f  0*0  >  aDCl  so  on  for  the  successive  derived 
curves. 
Drawing  the  graphs  of  y  =  f  (x)  =  ^  and  y  =  /'"  (x)  =  2 

together  with  those  of  y  =  f  (x)  =  x2  and  y  =  J"  (x)  =  2  x 
(shown  in  Examples  3  and  4,  Art.  115),  it  is  seen  that  the 
areas  are  represented  as  stated. 

It  may  be  seen  also  that    j  f(x)dx=   j  x2dx  =  -~  =  A, 

being  represented  by  the  ordinate  of  y  =  — ,  is  an  integral 

function  of  x2  and  the  graph  an  integral  curve  of  x2. 

If  y  =  2  be  the  fundamental  curve,  then  y  =  2  x  is  the 

xz 
first  integral  curve;  y  =  x2,  the  second;   y  =  -5-,  the  third; 

x* 
y  —  — ,  the  fourth;  and  so  on. 


CHAPTER  III. 

INTEGRAL  CURVES.     LENGTH  OF  CURVES.     CURVE 
OF  A  FLEXIBLE  CORD. 

140.  Integral  Curves.  —  If  F  (x)  has  /  (x)  for  its  deriva- 
tive, then  F(x)  is  called  an  Integral  Function  or  simply  an 

Integral  of  fix).     The  General  Integral  is    /  /  (x)  dx  =  F  (x) 

+  C,  called  also  the  Indefinite  Integral. 

The  graph  of  an  integral  function  is  called  an  integral 
curve.     If  the  original  or  fundamental  function  is 

y=f(x),  (i) 

then  y  =  F(x)  (2) 

is  the  first  integral  curve  of  the  curve  (1),  where  F  (x)  is  that 
integral  of  /  (x)  which  is  zero  when  x  is  zero.  In  the  general 
figure  of  Art.  125,  F  (x)  is  the  area  OPM  under  the  curve 
y  =  f  (x);  in  the  figure  of  Art.  139,  if  y  =  /  (x)  =  x2,  then 
F(x)  is  the  area  under  y  =  x2  and  is  the  ordinate  of  the 

integral  curve  y  =  — 
o 

It  is  manifest  that  for  the  same  abscissa  x,  the  number 
that  indicates  the  length  of  the  ordinate  of  the  first  integral 
curve  is  the  same  as  the  number  that  represents  the  area 
between  the  original  curve,  the  axis  (or  axes  for  some  func- 
tions), and  the  ordinate  for  this  same  abscissa.  Hence,  the 
ordinates  of  the  first  integral  curve  may  represent  the  areas 
of  the  original  curve  bounded  as  stated. 

It  may  be  seen  also  that  for  the  same  abscissa  x,  the  number 
that  expresses  the  slope  of  the  first  integral  curve  is  the  same 
as  the  number  that  measures  the  length  of  the  ordinate  of 

227 


228 


INTEGRAL  CALCULUS 


the  original  curve.  Hence  the  ordinates  of  the  original 
curve  may  represent  the  slopes  of  the  first  integral  curve. 

The  integral  curve  of  the  curve  of  equation  (2)  is  called 
the  second  integral  curve  of  the  original  curve  of  equation 
(1).  The  integral  curve  of  the  second  is  called  the  third 
integral  curve  of  the  original  curve  (1),  and  so  on.  Thus 
for  any  given  curve  there  is  a  series  of  successive  integral 
curves.* 

The  function  cos  0  and  the  first  and  second  integral  curves 
are  shown  with  their  graphs. 


Jt/2  7t  3Jy2  2Jt 

Let  y  =  cos  8  be  the  fundamental  function,  then 

y  =   f  cos  6  dd  =  sin  0  +  C, 

where  C  is  zero,  as  y  is  zero  when  0  is  zero;  and 

y  =    I  sinddd  =  -cos0  +  C, 

*  The  statements  in  the  three  paragraphs  above  with  some  difference 
of  words  are  givon  in  Murray's  Integral  Calculus,  where  a  fuller  treat- 
ment will  be  found  in  the  Appendix. 


APPLICATION  TO  BEAMS  229 

where  C  is  one,  as  y  is  zero  when  0  is  zero.     Hence, 

y  =  sin  0     and     y  =  1  —  cos  0  =  vers  0 

are  the  first  and  second  integral  curves  of  the  curve  z/  =  cos0. 

It  is  seen  that  the  ordinate  of  the  first  integral  curve  at 
$  =  x/2  is  +1,  that  number  being  the  same  number  that 
measures  the  area  under  the  fundamental  curve  for  the  same 
abscissa;  the  ordinate  being  zero  at  &  =  it  indicates  that  the 
algebraic  sum  of  the  areas  of  the  fundamental  curve  is  zero 
and  hence  that  the  area  below  the  axis  from  0  =  x/2  to  t  is 
exactly  equal  to  that  above  from  0  =  0  to  x/2;  the  ordinate 
being  zero  again  at  0  =  2  x  indicates  that  the  areas  of  the 
fundamental  curve  above  and  below  the  axis  are  exactly 
equal  up  to  0  =  2w. 

It  is  manifest  that  the  ordinates  of  the  second  integral 
curve  indicate  the  corresponding  areas  for  the  first  integral 
curve,  the  number  being  +2  from  0  =  0  to  x  and  2  —  2  =  0 
up  to  0  =  2  w. 

In  the  case  of  this  function  the  series  of  integral  curves 
can  be  extended  indefinitely  without  any  difficulty. 

It  is  manifest  also  that  the  ordinate  at  any  point  on  the 
fundamental  curve  gives  the  slope  at  the  corresponding 
point  on  the  first  integral  curve,  the  ordinate  for  the  first 
gives  the  slope  of  the  second,  and  so  on. 

The  subject  of  successive  integral  curves  has  useful  appli- 
cation to  problems  in  mechanics  and  engineering. 

Illustrative  examples  follow,  showing  the  application  to 
the  expression  and  graphical  representation  of  the  shearing 
force  and  bending  moment  throughout  the  length  of  a  loaded 
beam;  also  to  the  slope  and  deflection  of  the  elastic  curve, 
the  curve  of  the  mean  fiber  of  the  material  of  the  beam. 

141.  Application  to  Beams.  —  As  given  in  the  Mechanics 
of  Beams,  the  Vertical  Shear  at  any  section  of  a  loaded  beam 
is  the  algebraic  sum  of  the  vertical  external  forces  on  either 
side  of  the  section,  and  the  Bending  Moment  is  the  algebraic 


230  INTEGRAL  CALCULUS 

sum  of  the  moments  of  those  forces  about  a  point  in  the 
section.  The  moment  of  a  force  about  a  point  is  the  product 
of  the  force  and  the  length  of  the  perpendicular  from  the 
point  to  the  line  of  action  of  the  force.     (See  Art.  172.) 

The  Elastic  Curve  is  the  curve  assumed  by  the  mean  fiber 
along  the  axis  of  a  longitudinal  section  through  the  centers 
of  gravity  of  the  cross  sections  of  the  beam,  the  Slope  is  the 
slope  of  the  tangent  to  the  curve  at  any  point,  and  the 
Deflection  is  the  ordinate  at  that  point. 

Taking  abscissas  to  denote  as  usual  horizontal  lengths, 
the  ordinates  will  represent  the  quantities  to  be  depicted  by 
the  curves.  The  fundamental  curve  is  the  curve  for  L,  the 
load;  the  Shear  S  is  represented  by  the  first  integral  curve; 
the  Moment  M  by  the  second  integral  curve;  EI  upon  the 
Slope  m,  by  the  third  integral  curve;  and  EI  upon  the  De- 
flection d,  by  the  fourth  integral  curve,  where  E  and  I  are 
constants  denoting  the  Modulus  of  Elasticity  and  Moment 
of  Inertia,  respectively. 

Example  1.  —  Let  a  beam  of  length  I  between  supports 
be  simply  supported  at  each  end  and  loaded  with  a  uniform 
load  of  w  lbs.  per  linear  ft. 

I,  =  y  =  —w,  w  taken  with  negative  sign  as  a  downward 
force. 

S  =  y  =    I  —wdx  =  —wx 

+  [So  =  -w,S  being  zero  when  x  =  -A  • 

M  =  y  =  /(t  "  wx) dx  =  y x  ~  if  +  (Mo  =  °' M  when 

x  =  0). 

„T/  s        C  (wl        wx2\  ,        wl  2      wx3 

EI(m  -If)  -  J  (y *  "  -2)dx  =  T*  "  X 

-j-  ( mo  =  —  —  wl3,  m  when  x  =  0,  m  being  zero  at  x  =  ^  J  • 


EI 


APPLICATION  TO  BEAMS 
dx 


231 


r/wlx2      ivx*      wl*\ 


Wlxz        IVX*        wl3X    .     ,  ,  nix  n\ 

~12"  ~  2T  ~  "2T  +  (  °  =    '  )# 


EXAMPLE  1. 


Load  Line 


7     5    wl^m„^ 

dosSSIrlrmtaR 


- 


EXAMPLE  2 


*==ZZ1 


Inflexion  Point      d=-j^^,max.      Inflexion  Point 


—  ■  .,.,,.,,,;  n 


XDe-flection 
Curue 


^Deflection 
Curve 


*-0grfJ&-3f ~-~y3V3l=  0.58-1 -A^^^-J 


XSlope 
Curue 


M=2/24  wl?max. 
M=-  2/J2  wl2,max.  tieg. 
Shear  and  Load  Lines  the  same  as  Example!. 


XMoment 
Curve 


232  INTEGRAL  CALCULUS 

Example  2.  —  Let  a  beam  of  length  I  between  supports  be 
fixed  at  each  end  and  loaded  with  a  uniform  load  of  w  lbs. 
per  linear  ft. 

L  =  y  =  —w. 

S  =  y  =       —  wdx=—wx 

+  [So  =  ~k  f  S  being  zero  when x  =  ■=  J- 

M  =  y  =  J  {2~wx)dx  =  jx~^ 

wl2 
+  (M0.  M  at  x  =  0),  Mo  =  -  -yz ,  see  below. 

EJ  (m  -  ?/)  =  J  [^-x  -  -g-  +  Moj  dx  =  -j-z2  -  — 
+  MqX  +  (m0  =  0,  m  at  z  =  0) 

on7  1^3^  /Wl2  \ 

=  -j-  x2 ^ —  (  jo  x  =  —  M0x  j ,  from  m  =  0  at  x  =  Z. 

nr  /j  \  C  (Wl     i        WX*        wl2     \  J 

»tf-iF)-J(T^--r"i27<fa 

=  "12"  ~  24"  ~~  ~24~  +  (*>  =  °>  d  at  x  =  0). 

Example  3.  —  Cantilever  Bridge.*  In  the  cantilever 
bridge  the  joints  are  placed  at  the  inflexion  points,  where 
the  Bending  Moment  would  be  zero  if  the  bridge  were  con- 
tinuous over  the  whole  span. 

Let  the  beam  of  Example  2  have  joints  at  the  inflexion 
points,  and  let  the  length  of  each  cantilever  arm  be  denoted 
by  a  and  the  length  of  the  suspended  span  by  6. 

The  Shear  line  and  the  Moment  curve  will  be  unchanged 
but  the  Slope  and  Deflection  curves  will  not  be  continuous  as 
they  were  without  joints  in  the  beam.     For  slope  and  de- 

*  The  essential  features  of  this  example  are  given  by  H.  E.  Smith 
in  his  "  Strength  of  Material." 


APPLICATION  TO   BEAMS 


233 


flection  each  part  is  to  be  considered  as  a  separate  beam,  the 
arms  having  in  addition  to  their  uniform  load  a  concentrated 
load  at  their  jointed  ends,  equal  to  half  the  uniform  load  on 
the  middle  span  b,  since  that  part  is  supported  at  the  joints 
by  the  arms. 


L  =  —  w. 


S  being  zero  when  x  =  ~  r 
=   f(^(2a  +  b)~wx)dx  =  ^(2a  +  b)x- 

+  (m0=  -™(a  +  b)\    M  =  0atx  =  a. 


-^               ,,      n      w  fe%      .  ,v         vox1      wa  ,        , . 
From        M  =  0  =  ^  (2a  +  b)  x  -  ^ g-  (a  +  &), 


M 


wrr 


a:  =  a    and    rr  =  a  +  6, 


that  is, 


x=(l-iv/3)!/2    and    a;  =  (l+  i V® 2/2, 
and 
&  =  1  V3  J,   from   M  =  0  =  -^ 2~~T2       ExamPle  2* 


234  INTEGRAL  CALCULUS 

It  is  seen  that  the  maximum  negative  moment  TV  wl2  at 
the  fixed  ends  is  twice  the  maximum  positive  moment  ^  wl2 
at  the  middle.  For  equal  strength,  the  bending  moments  at 
the  ends  and  middle  of  the  beam  should  have  equal  numeri- 
cal values;  this  requires  that 

wa  ,     .  ,N      wb2 
T(a  +  6)  =  -g-, 

that  is, 

4a2-f46a  =  62;    or    4a2  +  46a  +  b2  =  2b2; 
or  (2a  +  6)2  =  262, 

/.     b  =  i  V21     and     a  =  (l  -  J  V2)  1/2. 
It  follows  that 

wb2      wl2  ,         wa ,     ,  , N  wl2 

__  =  __   and  __(a+6)  =  __, 

giving  equal  strength  at  ends  and  middle  of  beam. 

Example  4.  —  Let  a  beam  of  length  I  between  supports  be 

fixed  at  the  right  end  and  simply  supported  at  the  left  end; 

and  let  it  be  loaded  with  a  load  uniformly  increasing  from 

zero  at  the  left  end  to  w  lbs.  per  linear  ft.  at  the  right  end. 

T  w 

L  =  y=    -jx. 

/w  w 

—  jxdx  =  —  tyj x2  +  (£0  =  S,  when  x  =  0). 

M  =  y  =  f(s0  -  ~x2^j  dx  =  Sox  -  ^x* 
+  (Mo  =  0,  M  when  x  =  0). 

EI  {m  _  y)  _  j(SfrT  _ • *yx  =  *£  -  £, 

+  (mo=o7  —  ^- ,  m  at  x  =  0  J ,  from  m  =  0  at  a:  =  Z. 
ett  /j        \         T/rf      S0l2  .  S0X2        w     A  , 

rf        fiW2        aSox3       wxb       ,  ,  ,    .  . 


APPLICATION  TO  BEAMS 


235 


__.,        .      ft      wl*      SQl*      Sol*       wP       ,  , 

EI  (d  =  y)  =  0  -  24  -  T  +  T  "  S511  when  *  =  *' 

.*.     £0  =  77y  supporting  force  at  0. 

0  w£       it;    „~|  4     7 

•'•    s  =  »  =  io-2ia;2L=-iowZ' 

4 
Si  at  right  end;  supporting  force  =-  + —  wl. 


°r°mmmm^^ 


Load  Line 


^liiiiiiimiiijiiiiin 


V-—ysVsl=0.447l 


^^^^miiiiii^ 


Shear  Curoe 


M=, 0.03 >wl? max. 


■TTTTTrmTT^ 


at  right  end, M=  0. 06 z/5  wl2,  max.  nee/. 


'.  nea.  ^\ 


^TT-mnTTTTTT 


HI 


VsVmsI  =0.77-1 


^uiff 


ppp^— 


Moment 
Curoe 

X 

Slope  Carve 


XDeflection 
Curve 


0.916  wl^ 
384  EI 


^      U       10      2^  ' 

x  =  i  a/5  Z  =  0.447  .  .  .  I,  when  the  shear  is  zero. 

, ,  wl         wx3~\  v  5    10      A  no    79 

M  =  y  =  iox  -  eiL^r  n*  =  003 wP> 


236  INTEGRAL  CALCULUS 

maximum  positive  moment  where  shear  is  zero,  since 

ax 

M-y  =  rox-t\,r-kwl2=-°Miwl2' 

maximum  negative  moment. 

,,      _      wl        wx3 

M  =  0=idx-JT> 

.-.     x  =  Vfl  ^  i  V3X  5  =  0.77  , .-.  .  I , 
where  moment  is  zero  and  where  there  is~an  inflexion  point 
on  the  Elastic  Curve,  the  fourth  integral  curve,  since  M  is 
second  derivative  of  d  with  respect  to  x. 

„T/  N  wl3     .     Wlx2        WX4~]  wl3      r.r  . 

EI  (m  =  y)  =  -  —  +—  -  —^=  -—,E1  upon  slope 

at  left  end. 
n  _    _  wl3      wlx2  _  wx* 
~  120  +  "20"      Wl' 
.'.     x  =  £  V  5  I  and  I,  when  slope  is  zero. 
,-,T  , .        N  wl3  wlx3      'wxbl  0.016    r=    71 

*nd=v)  =  -mx+w-miU«i=-^-VEv* 

=  —0.002385  wl4  =  — ^-r-wl4, EI  upon  max.  deflection. 

The  deflection  is  zero  at  x  =  0  and  x  =  I,  as  used  above. 

142.  Lengths  of  Curves.  —  Rectangular  coordinates.  —  It 
has  been  shown  in  Art.  10,  (d)  that  ds2  =  dx2  +  dy2;  now  let  s 
denote  the  length  of  the  arc  whose  ends  are  the  points 
(xo,  2/0)  and  (x,  y),  then  ds  =  Vdx2  +  dy2;  whence 

•=£[■+ ($*]'-■ 

or 

according  as  ds  is  expressed  in  terms  of  x  or  of  y. 


LENGTHS  OF  CURVES  237 

In  getting  the  length  of  any  curve,  that  formula  which 
gives  the  simpler  expression  to  integrate  is  the  preferable  one 
to  use. 

EXERCISE   XXVm. 

1.  Find  s  of  the  circle  x1  +  y-  =  a2,  and  the  circumference.     Here 

dy  _  _  x .    /dy  V  =  z2  , 
dx  y'    \dxj       y2' 

hence  s=   PTl  +  ^~P dx  =    f *&  +  *)* dXf     using  (1), 

J"x       dx                 .     .X~]x            .     .  x  .     ,xo 

.  =  a  sin  x-       =  a  sin-1 a  sin-1  — 

x0  Va2  —  x2  °-Jxo  «  a 

Circumference,  s  =  4  a  sin-1  -      =  4  -  a  =  2  7ra. 
aJo         2 

2.  Find  s  of  the  semi-cubical  parabola  ay2  =  x3.    Here 

fdyY  =  9xm 
\dxj       4a' 

"""     S  =  J    V1  +  Ta)dx'     "^g  (1)> 

From  the  origin,  s  =  ^  [(*  +  4^)*  ~  x]  ■ 

3.  Find  s  of  the  cycloid  x  =  a  arc  vers  -  =F  V2  ay  —  y2.     Here 

/<faV  =     y     . 

\d?//        2a-2/' 

.-.      s  =  VTa  P(2a  -  y)~hdy,    using  (2), 

=  2V2"a[(2a-2/o)'-(2a-^]. 

Making  y0  =  0  and  y  =  2  a  and  taking  twice  the  result  gives  8  a  for 
the  length  of  one  arch.     (See  Ex.  3,  Art.  97.) 

Note.  —  Finding  the  length  of  a  curve  is  called  rectifying  the  curve, 
since  it  is  getting  a  straight  line  of  the  same  length  as  the  curve.  The 
semi-cubical  parabola  was  rectified  by  William  Neil  and  by  Van  Heu- 
raet  also,  and  it  is  the  first  curve  that  was  absolutely  rectified.  The 
second  rectification,  that  of  the  cycloid,  was  by  Sir  Christopher  Wren 
and  by  Fermat  also,  and  the  third  was  of  the  cissoid  by  Huygens. 
These  rectifications  were  effected  before  the  development  of  the 
Calculus. 


238  INTEGRAL  CALCULUS 

4.   Find  s  of  the  parabola  x2  =  2  py,  (x0,  yo)  being  the  origin.     Here 

fdy\  =  t. 
\dx)       v2 ' 

•■•  «-n*+ 3** -if  <*+.*»** 

=  -  [|  Vp^  +x2  +  ^log (z  +  vy +z!)]J  [Ex. 7,Art.  123.] 

_  X 

ea  +  e   a  J .     Here 

=X[Ke°+2+e  aXr* 


=  X   2r  +  e    a)dx=2\ea-e    V 


6.   Find  s  of  the  ellipse  y1  =  (1  -  e2)  (a2  -  x2),  and  the  length  of 
the  curve,  e  being  the  eccentricity.     Here 
dy  _  -_(1  -  erfx 
dx         \/'c 


1  -  erfx  .       /^  V  =  (1  ~  e2)  *2  . 
'-tf^tf     '       \dx)  =      a2-x2    ' 

C  {a2  -  e2x2)- -M=;  (1) 

J  Xn  Vrt2    _    7-2 


hence,  the  length  of  the  elliptic  quadrant  sg  is 

sq  =  f  (a2  -  c2*2)*  -7=='  (2) 

^o  Va2  —  x2 

For  the  integration  of  (1)  and  (2),  see  Ex.  5,  Art.  203  and  Note. 

7.    Finds  of  the  circle,  and  the  circumference,  using  the  parametric 
equations  x  =  a  cos  0,  y  =  a  sin  0.    Here 


fdx\      sin2  0 

IT"  I   =  ^.-^  5     dy  =  a  cos  Odd; 


cos20 


J~0  H2ir 

d0  =  a  (0  -  0u)         =2  ?ra. 
0U  Jo 


LENGTHS  OF  POLAR   CURVES  239 

More  directly; 

pd  "l2ir 

ds  =  add;     :.     s  =  a\    dd  =  a  (0  -  0O)        =  2ira. 
Jvo  Jo 

8.  Find  s  of  the  involute  of  the  circle,  and  length  of  the  arc  between 
0=0  and  0  =  r,  using  the  equations 

x  =  a  (cos  0  +  0  sin  0),     y  =a  (sin0  —  0  cos0). 
(See  Cor.  Ex.  2,  Art.  97.)  Ans.  \  ad2;  \  air2. 

9.  Find's  of  the  cycloid,  and  the  length  of  one  arch,  using  the  para- 
metric equations  x  =  a  (0  —  sin  0),  y  =  a  (1  —  cos  0).     Here 

dx  =  a  (1  —  cos  0)  dd;     dy  =  a  sin  0  dd; 

:.     s  =   f   (dx2  +d?/2)*  =  a  V2   f  (1  -  cos  0)*c?0 
=  2  a  I    sin  -  dd,  since  Vl  —  cos  0  =  V2  sin  - 

=  _4aC032j9.  =  4alC°S2-COS2JJo    ; 

?  =  8  a,  for  one  arch. 

10.  Find  s  of  the  parabola  y2  =  2  px,  (x0,  t/o)  being  the  origin,'  and 
the  length  of  the  arc  from  the  vertex  to  the  end  of  the  latus  rectum. 

l  =  p/2[V2  +  log(l  +  V2)]. 

11.  (a)  Find  s  of  the  hypocycloid  x*  +  y*  =  a*,  and  the  length  of 
the  curve.  (a)  Ans.  §a*  (x*  -  xo§);  6  a. 


(6)  Whenx  =  asin30,  y  =  a  cos3  6. 


ff  a  sin2  0l 


(6)  4ns.  4    fa  sin2  0       =  6  a 


143.  Lengths  of  Polar  Curves.  —  It  has  been  shown  in 
Art.  77,  (3),  that  for  a  polar  curve  ds2  =  P2dB2  +  dp2.  Now 
let  s  denote  the  length  of  the  arc  whose  ends  are  (p0,  0O)  and 
(p,  0) ,  then  <is  =  Vp2  d02  +  <jp2 ;  whence 

s=    feVP*d0*  +  dp*     or     s=    f  Vp2dd2+dp2,     (1) 

according  as  ds  is  expressed  in  terms  of  6  or  of  p. 

If  the  formulas  of  the  preceding  Article  for  lengths  of 
curves  be  transformed  to  polar  coordinates  by  making 
x  =  p  cos  6  and  y  =  p  sin  0,  (1)  of  this  Article  results. 


240  INTEGRAL  CALCULUS 

EXERCISE   XXIX. 

1.  (a)  Find  s  of  the  circle  p  =  2  a  cos  0,  and  the  circumference. 

s  =    C  V4  a2  cos2  0  +  4  a2  sin2  0  d0  =  2  a  L  d0 

=  2a(0  -0o)T  =  27ra. 
(6)   For  the  circle  p  =  a; 

s  =    (     p  dd  =  a  (     d0  =  a  (0  -  0o)         =  2ira. 
Je0  Je0  Jo 

2.  Find  s  of  the  cardioid  p  =  2  a  (1  -  cos  0),   and  total  length. 
Here  dp  =  2  a  sin  Odd; 

...     s  =    C6  [4  a2  (1  -  cos  0)2  +  4  a2  sin2  0]*  dd 
J  d0 

=  2  a  f    V2  (1  -  cos  6)*dd  =  4  a  f   sin  -  d0 
Joo  ^00        2 

0    r     0o         0"|27r     1A 

=  8  a    cos  ^  -  cos  o         =  16  a. 

3.  Find  s  of  the  spiral  of  Archimedes  p  =  ad,  and  the  length  of  the 
first  spire.     Here 

ds  =  Vatd2  +  a2  dd  =  a  Vl  +  02d0; 

...     s  =  a   f VrT0~2  d0  =  ^  fa  ^1+"^  +  log  (0  +  V1  +  0OT 

»/0o  ^  L  J0O=° 

=  a  W  Vi+4tt2  +  Uog  (2tt  +  Vl  +4tt2)]. 

4.  Find  s  of  the  logarithmic  spiral  p  =  c00  from  po  to  p,  and  the  length 
from  the  pole  (p  =  0)  to  p  =  1. 

00=  log  p;      /.     add  =  —      or     pdB  = — ; 
p  <J 


Jpo  V  a2 


2   ,    ,  ,       Vl  -f  a2 


a  Jo  « 

6.   Find  s  of  the  conchoid  p  =  a  sec  0,  and  the  length  of  the  arc  from 
0  =  0  to  0  =  tt/4.     Here 

dp  =  asec0tan0d0; 

...     s  =   (°  Va2  sec2  0  -f-  a2  sec2  0  tan2  0  d0  =  a  f   sec2  0  d0 
Jdo  Jo* 

—  a  tan  0       =  a  (tan  0  —  tan  0O)       =  a. 

Jflo  J  0 


CURVE  OF  A  CORD 


241 


144.  Curve  of  a  Cord  under  Uniform  Horizontal  Load  — 
Parabola.  —  When  a  flexible  cord  supports  a  load  which  is 
uniformly  distributed  over  the  horizontal  projection  of  the 
cord,  as,  for  example,  the  cable  of  a  suspension  bridge, 
which  supports  a  load  distributed  uniformly  per  foot  of 
roadway,  the  curve  assumed 
by  the  cord  is  a  parabola. 
This  is  evident  geometri- 
cally, and  the  curve  can  be 
shown  analytically  to  be  a 
parabola. 

Consider  a  portion  OP  of 
the  cord  AOB,  0  being  the 
lowest  point  of  the  cord. 
The  equilibrating  forces 
acting  on  OP  are  the  ten- 
sions H  and  T  at  the  ends, 

and  the  weight  of  the  load,  acting  at  the  center  of  OM. 
Since  three  forces  to  be  in  equilibrium  must  meet  in  a  point, 
the  tangent  to  the  cord  at  P  passes  through  the  middle  of 
OM.  Now  it  is  a  property  of  the  parabola  that  the  sub- 
tangent  is  bisected  at  the  vertex,  in  which  case  OC  =  J 
NP  =  CM.  Hence  the  curve  assumed  by  the  cord  is  a 
parabola. 

Otherwise  the  analytic  conditions  of  equilibrium  give 


%MP*  = 


VOX 


X 


i-fl»-0! 


w 


(1) 


the  equation  of  a  parabola.  The  same  result  is  gotten  by 
taking  the  sides  of  ^he  triangle  PDT  to  represent  the  three 
forces  T,  H,  and  wx,  and  then, 

J(*x  wx  dx      "'^2 
o    S- 


dy 
dx 


wx 

H 


WX' 

2H 


(1) 


Algebraic  sum  of  moments  of  forces  about  point  P. 


242  INTEGRAL  CALCULUS 

T  cos  <f>  =  H  and  T7  sin  0  =  wx  give  by  division 

,       .       wx      dy         ,    , 
tan 4>  =  -rr  =  ~r,  as  before. 
H      ax 

TJ  

Also.  T  — =  VH2  +  (wx)2  gives  the  tension  at  any 

1  COS  <p  '    v      /      B  J 

point  of  the  cord. 

The  curve  is  thus  shown  to  be  a  parabola  with  its  vertex 
at  0  and  its  axis  vertical.  If  the  supports  at  A  and  B  are  at 
the  same  elevation  and  I  is  the  span,  the  sag  d  at  0  is  y,  for 

X  =    2  '  J 

«*  (2) 

(3) 


d  = 

$H' 

H  = 

wl2 
'8d' 

and     Ti 

dy  = 
dx 

-  tan  4>\ 

wl 
~2H~ 

COS  0i 

where  <fo  is  the  angle  of  inclination  of  PT  at  the  supports. 
The  length  of  the  cord  for  a  given  span  and  sag  is  gotten  by 


(5) 


Let  -rr  —  ~i  or   —  =  a,  to  simplify:  then, 
Ha  w 

.favy^+jic(i±^5?). 

Replacing  tt  =  - ,  s         =  Si 
/i       a      J.r=o 


CURVE  OF  A  CORD  243 

Total  length  =  2  05  =  2«i 

T/Z  IE*      V\  ml 

(7) 


wi   Zip    p    h    r/i    ./F~~p\«,-| 


Corollary.  —  Expanding  the  two  terms  of  (7)  into  series 
(as  shown  later)  and  adding  like  terms,  the  total  length  of 
the  parabola  in  terms  of  I,  w,  and  H  is 

rn   x    i  i  +u  7    1       W2Z3  W4Z5         .  W6?7  /0, 

Totallength  =  /;  +  — -^^  +  _^ ,     (8) 

72 

or  in  terms  of  I  and  d,  since  #  =  ^-^  from  (3), 

t  +  n       +u       7   .   8c?2       32^4   i   256^6  ft* 

Total  length  =  ?,  +  _-—  +  _ ...  (9) 

Also,  expanding  (5)  by  the  binomial  theorem  and  integrating 
gives 

s  =  -   f'a2  +  tf)idx 

a  Jo 

=  -  fX\a+£--  E^i  +  7^ ]dx       (x2  <  a2) 

a  Jo  |_        2  a      8  a3      16  a0  J  v  ' 

ir    I  ^     x5       x7 

~a|_a        6a      W  +  112d6       '"'J 

X3  X5  X7  ,  1  _   w 

~X^~§oJ      W  +  112a1       '  '  "  '  wnere  a~H' 

/.      Total  length,  5  =  2  Sl 

=  ,        ufl8  i^Z5  i^7      _  fm 

"t'24i72      640  #4_t~  7168  #6  '  '         w 

The  total  length  can  be  found  by  (8)  or  (9)  to  any  desired 
degree  of  accuracy  and  it  can  be  gotten  exactly  by  (7) ;  when 
d  is  quite  small  compared  with  I,  then  the  third  and  succeed- 
ing terms  in  (8)  and  (9)  are  so  small  that  they  may  be  neg- 
lected giving: 

wHz  8  d2 

Total  length,  S  =  I  +  Krm  =  &  +  ^T>  approximately.     (10) 


244 


INTEGRAL  CALCULUS 


When  S  and  H  are  given  to  find  I,  a  cubic  equation  results, 

the  solving  of  which  may  be  avoided  by  putting  *S3  for  P, 

since  they  are  nearly  equal  when  d  is  small;  then, 

w2S3 
1  =  S  -  247/2  >  approximately.  (11) 

145.  The  Suspension  Bridge.  —  The  cables  of  a  suspen- 
sion bridge  are  loaded  approximately  uniformly  horizontally, 
since  the  roadway  is  horizontal,  or  nearly  so;  and  the  extra 
weight  of  the  cables  and  the  hangers  near  the  supports  is  a 


small  part  of  the  total  load  carried  by  the  cables.     As  shown 
in  Art.  144,  under  the  conditions  stated,  the  curve  of  the 

WX" 

cables  is  the  parabola  whose  equation  is  y  =  ^tj,  where  H 

is  the  horizontal  tension. 

Example  1.  —  The  Brooklyn  Bridge  is  a  suspension  bridge 
which  has  stays  and  stiffening  trusses  to  prevent  oscillation. 
The  span  between  main  towers  is  1595  feet. 

If  the  sag  OD  is  128  feet  and  the  weight  per  foot  supported 
by  each  cable  is  1200  lbs.,  without  considering  the  stays  or 
stiffening  trusses,  to  find  the  terminal  and  horizontal  ten- 
sions, substitute  the  numerical  values  in  (3)  of  Art.  144.  For 
terminal  tension, 


Ti  =  H  sec  0! 


or 


Tx 


VWU+- 


For  horizontal  tension, 


H  =  |^  =  2,981, 279  lbs. 


Ti  =  H  sec  0i 


id 
Sd 


VP  +  1Q<P  =  3,128,929  lbs. 


CURVE  OF  A  FLEXIBLE  CORD  —  CATENARY       245 

Example  2.  —  A  eord  loaded  uniformly  horizontally  with 
1  lb.  per  foot  is  suspended  from  two  supports  at  same  eleva- 
tion and  200  feet  apart  with  a  sag  at  middle  of  50  feet.  Find 
the  length  of  the  cord. 

Here  tan^i  is  equal  to  f{|  =  1; 

wl       200  w      i 
...     tan01  =  — =  — =1; 

■'•    ^  =  2oo  or  nrm*m. 

w     2  _  JL_   2 
y~2HX~      200*/ 

P  I5 

S  =  total  length  =  I  +  ^^  -  64Q^4  +  "  ". " 

-  9004-     (20Q)3  (2QQ)5     +  ...  -  228  TO  annrox 

"  2°°+  24(100?  "  640000?  +  '    PP 


-V(iro,.  +  (W  +  ,Mlo8[1M  +  ^"«"'] 

=  100  V2  +  100  log  [1  +  V2]  =  141.42  +  88.14 
=  229.56,  exactly. 

146.  Curve  of  a  Flexible  Cord  —  Catenary.  —  If  a  per- 
fectly flexible  inextensible  cord  of  uniform  density  and  cross 
section  be  suspended  from  two  fixed  points  A  and  B,  it 
assumes  a  position  of  equilibrium  under  the  action  of  gravity. 
The  curve  thus  formed  is  the  Catenary,  whether  the  line 
joining  the  points  of  support  is  horizontal  or  not. 

To  find  the  equation  of  the  curve  AOB;  let  w  denote  the 
weight  of  a  unit  length  of  the  cord  and  s  the  length  of  the 
arc  whose  ends  are  the  lowest  point  (0,  0)  and  the  point 
(x>  y),  or  P;  then  ws,  the  weight  of  the  arc  OP,  is  in  equi- 
librium with  the  tangential  tensions  at  0  and  P.     Denote 


246 


INTEGRAL  CALCULUS 


the  horizontal  tension,  which  is  the  same  at  all  points,  by 
H—wa.  If  c  •  PT  represents  the  tangential  tension  T  at  P, 
c  •  PD  and  c  •  DT  will  represent,  respectively,  the  horizontal 
and  vertical  tension  at  P. 


Hence, 


v+- 

—  > 

r 

fi~ 

-iC 

■AK- Xz--- 

---> 

< 

N 

pf  'u 

\Z 

r 

& 

ysj> 

2/z\    n 

0 

— [~- X 

a 

T/< 

ITS 

ifl 

F 

i 

Mz 

°i 

i» 

r    «i    - 

dy  _  C'  DT  _ws  _s 
dz      c  •  PD      wa      a ' 


s  _  v  ds2  -  dx1 
a 


dx 


dx 
a 


ds 


Va2  +  s2 ' 
+  Va2  +  s2" 


«_  p    *      ,  r.+v*+*i 

a      Jo  Va2  +  s2  L  a  J 


t2  +  52 

The  exponential  equation  is 


ea  = 


+  Va2  + 


ac" 


+  V^+72, 


which  solved  for  s  gives  for  length  of  OP, 

S  =  |(^-e")=asinh5. 
(Compare  Ex.  5,  Art.  142.) 


(1) 


(2) 


(3) 


CURVE  OF  A  FLEXIBLE  CORD  —  CATENARY       247 

dii 
Substituting  in  (3),  s  =  a -p.  from  (1),  gives 


'hi 

dx 


if  -      --\  x 

=  2Y~e     I  =sinha'  (4) 

•••  J!* = l£(e*  -  •"■)* = I  sinh^"; 

2/  +  a  =  |  f  e~a  +  e   aJ  =  a  cosh-  (5) 

is  the  equation  of  the  catenary  referred  to  the  axes  OX,  OY. 
If  the  origin  of  coordinates  is  taken  at  a  distance  a  below 
the  lowest  point  of  the  curve  and  the  curve  be  referred  to 
the  axes  O1X1  and  OiY,  its  equation  is 

y  =  a  cosh-  =  ^(e°  +  e   aj-  (5i) 

The  horizontal  line  through  Oi  is  called  the  directrix  of  the 
catenary,  and  Oi  is  called  the  origin. 

Corollary  I.  —  Since  a  =  OiO  is  the  length  of  the  cord 
whose  weight  is  equal  to  the  horizontal  tension,  and  there- 
fore the  tension  at  the  lowest  point  0,  it  follows  that  if 
the  part  AO  of  the  curve  were  removed  and  a  cord  of  length 
a,  and  of  the  same  weight  per  unit  length  as  the  cord  of 
the  curve,  were  joined  to  the  arc  OP  and  suspended  over  a 
smooth  peg  at  0,  the  curve  OPB  would  still  be  in  equihbrium. 

Corollary  II.  —  Since  the  sides  of  the  triangle  PTD  are 
proportional  and  parallel  to  the  three  forces  under  which 
the  arc  OP  is  in  equihbrium,  it  follows  that : 

tension  at  P  =  c-PT  __   T__d±_y 
tension  at  0  ~  c  •  PD  ~  wa  ~  dx~~  a' 

from  (3)  and  (5i),  by  differentiating  (3); 

-      T  =  wy; 

that  is,  the  tension  at  any  point  of  the  catenary  is  equal  to  the 


248  INTEGRAL  CALCULUS 

weight  of  a  portion  of  the  cord  whose  length  is  equal  to  the  ordi- 
nate at  that  point. 

Therefore,  if  a  cord  of  uniform  density  and  cross  section 
hangs  freely  over  any  two  smooth  pegs,  the  vertical  portions 
which  hang  over  the  pegs  must  each  terminate  on  the  direc- 
trix of  the  catenary. 

Corollary  III.  —  Subtracting  the  square  of  (3)  from  the 

square  of  (5i)  gives 

y2  =  s2  +  a2;  (6) 

dx2 
and  from  (6),  after  substituting  a2  =  y2^  from  Corollary  II, 

s  =  yTs-  (7) 

From  M,  the  foot  of  the  ordinate  at  P,  draw  the  perpendicu- 
lar MTU  then  PTi  =  y  cos  MPT^  =  y-£,  which  in  (7)  gives 
PTt  =  s  =  the  arc  OP;  (8) 

and  since  y2  -  P7Y  +  7W2,  from  (6)  and  (8), 

TiM  =  a.  (9) 

Therefore,  the  point  Ti  is  on  the  involute  of  the  catenary 
which  originates  from  the  curve  at  0;  7W  is  a  tangent  to 
this  involute;  and  TJ?,  the  tangent  to  the  catenary,  is 
normal  to  the  involute.  As  7W  is  the  tangent  to  this  last 
curve,  and  is  equal  to  he  constant  quantity  a,  the  involute 
is  the  equitangential  curve,  or  tractrix. 

147.   Expansion  of  cosh  x/a  and  sinh  x/a.  —  Expanding 

X  _x 

ef1  and  e  <*  in  series  and  taking  the  sum  of  the  two  series  term 
by  term  gives  for  (5i), 

=  °  +  fa  +  2&+---.  (10) 


APPROXIMATE  FORMULAS  249 

Taking  the  difference  of  the  two  series  gives  for  (3), 

s  =  a  sinh  -  =  a    -  +  -j^  +  -^i  +  ■  •  • 
a         \_a      a3  3!      a°o! 

.-.     for  (4), 

X 

For  these  expansions,  put  -  for  x  in  Examples  8  and  7  of 

Exercise  XLIII. 

148.   Approximate  Formulas.  —  The  series  in  Art.   147 

are  rapidly  convergent  when  x  is  small  compared  with  a. 

x 
Near  the  origin  -  is  a  small  quantity,  s  is  nearly  the  same  as 

CL 

sly  o  ■£ 

x}  and  -j-  =  -  =  - ;  hence,  neglecting  the  terms  with  the 
higher  powers  of  x ; 

y  =  a(l+^2\)  =  a+fa'    0I    y-a  =  fa'       (1) 

(X  x^   \ 

f  =  tan*  =  E  +  ~^7  -  -  +  A'  (3) 

dx  a      a3  3!      a      6  a3 

Equation  (1)  is  the  equation  of  a  parabola  with  its  vertex 
at  (0,  a),  the  lowest  point  of  the  catenary.  Where  the  cord 
is  very  taut  with  small  sag,  s  is  very  nearly  the  same  as  x, 
and  as  they  are  nearly  equal  in  any  case  near  the  vertex 

where  -^  is  small,  if  x  is  put  for  s  in  (1)  of  Art.  146,  then 
ax 


+  Wa2'  (2) 


the  equation  of  the  parabola  with  its  vertex  at  (0,0)  or  0. 


dy  _  x  m     .        _     C*  x  dx  _  x2 
dx~  a'  V  ~  Jo    ~a~  ~  2a 


250  INTEGRAL  CALCULUS 

Hence,  near  its  lowest  point  the  catenary  approximates 
in  shape  a  parabola.  When  xi  =  i  I,  y\  =  a  +  d,  d  denoting 
the  sag;  hence,  for  the  sag  at  the  lowest  point  of  the  curve, 

<*  =  £  +  »+-  -(from  (10),  Art.  147), 

or,  approximately, 

d  =  j-a  =  |J;  /.  H  =  g  [(2)  and  (3),  Art.  144].  (4) 
Also  at  the  supports  tan  <t>i  is,  approximately, 

tan^  =  ^  =  ^  =  ^    [(4),  Art.  144].  (5) 

When  Zi  =  o> 

Sl"2"t"48a2i"  "  "  ~2~t~48#2~1~'  *  '  ~2~1~6Z'i~'  "'  W 
Hence,  when  the  supports  are  at  the  same  elevation, 

73  7„273 

Total  length  =  2si  =  I  +  xts  =  1  + 


24a2  '  24#2 

=  I  +  ^,  approx.  [(10),  Art.  144].         (7) 

Note*  —  When  the  sag  is  1  per  cent  of  the  span,  the  error 
in  H  or  d,  from  using  these  approximate  formulas,  compared 
with  the  value  from  those  for  the  true  catenary,  is  about  ^? 
of  1  per  cent.  For  a  sag  of  10  per  cent  of  the  span,  the  error 
is  about  2  per  cent. 

149.  Solution  of  s  =  a  sinh  x/a.  —  While  in  practice  the 
approximate  parabolic  formulas  of  the  preceding  article  are 

x 
generally  used,  the  curve  y  =  a  cosh  -  appropriately  mag- 
nified fits  any  cord  hanging  under  its  own  weight,  the  con- 
stant a  depending  upon  the  tautness  of  the  cord. 

When  the  horizontal  distance  xh  from  the  lowest  point  of 

*  Poorman's  Applied  Mechanics. 


SOLUTION  FOR  THE   CATENARY  251 

the  curve  to  one  of  the  supports  B,  and  the  length  of  the  arc 
OB  =  si  are  given,  then  approximate  values  of  a,  yh  Th  H, 
and  d  =  yi  —  a  can  be  found.  For  putting  xi  for  x  in  (3), 
then 

Si  =  7i\ea—  e    a)  =  asinh—>  (1) 

2  a 

where  Si  and  X\  being  given,  a  is  found  by  solving  the  trans- 
cendental equation.  The  solution  is  by  approximations  and 
use  of  tables  of  the  hyperbolic  functions.  When  the  supports 
are  at  the  same  elevation  and  I  denotes  their  horizontal 
distance  apart  or  the  span,  I  =  2  xi}  and  for  total  length  of 
cord, 

£  =  2si  =  a\era-e~^)}  (2) 

where  S  and  I  being  given,  a  is  found  by  solving  the  equation. 
The  curve  heretofore  considered  is  called  the  Common 
Catenary,  the  cord  being  of  uniform  cross  section. 

In  the  Catenary  of  Uniform  Strength,  the  area  of  the  cross 
section  of  the  cord  at  any  point  is  varied  so  there  is  a  con- 
stant tension  per  unit  area  of  cross  section.  The  equation 
is 

y  =  clog  sec  (x/c). 

Example  1.  —  A  chain  62  feet  long,  weighing  20  lbs.  per 
ft.,  is  suspended  at  two  points  in  a  horizontal  line  50  ft.  apart. 
Find  a;  the  horizontal  tension;  the  terminal  tension;  the 
sag  of  the  chain. 

Here 

8  =  |(e«  -  e~*)  =  |(e«  -  e"°)  =  31.  (1) 

S  =  x  +  6^  =  25  +  ff~  =  31,    by  (2)'  Art'  148; 
(25)2  =  6X6 . 
"a2  25  ■  ' 

—  =  -  =  1.2,  approximately. 
a       o 


252  INTEGRAL  CALCULUS 

Now  let  —  =  2;     .-.     a  =  — ,  which  substituted  in  (1)  gives 

d  Z 

Let  /(2)  =  ez  —  e~z  —  2.48  2  =  0,  where  1.2  is  approx- 
imate value  for  z.  Reference  to  a  table  of  hyperbolic  sines 
will  show  that  the  value  of  z  is  between  1.1  and  1.2. 

The  following  is  a  method  of  getting  a  closer  approxima- 
tion to  the  value  of  the  root  z  without  the  use  of  tables.  Let 
z  =  Z\  +  h,  where  zx  is  an  approximate  root  differing  from 
the  root  z  by  a  small  quantity  h. 

By  Taylor's  Theorem : 

/(«)  -/(ft  +  h)  =  /(ft)  +  V  W  +  f /"to  + 0. 

Neglecting  higher  powers  of  h,f(zi)  +  hf  (ft)  =  0; 

.*.     z  =  ft  —  ^ .   ,  ■ ,  as  first  approximation  for  z. 
Let  this  value  be  22,  differing  from  2  by  k  <  h; 

■  ■   /(«)  =/fe  +  fc)  =/(*)+¥'<*)  +  ^/"(*)  +  •  •  •  =  0. 

Neglecting  higher  powers  of  k,  f  («2)  +  fc/'  fe)  =  0; 

as  second  approximate  value  for  2. 

By  repeating  this  process  a  value  which  approximates 
closer  and  closer  to  the  true  value  of  the  root  can  be  gotten. 
Applying  the  method  to  this  example : 

/to  e1-2  ~  e-1'2-  2.48  X  1.2  _      0.04  _      nnor 

/'to"  e1-2 +  e-^ -2.48  1.14"      UAWD' 

/.     Z2  =  ft  -  h  =  1.2  -  0.035  =  1.165. 


SOLUTION   FOR  THE  CATENARY  253 

Now 

_       gi-165  _  e-i.165  _  2.48  X  1.165  _  0.005 _  _nm- 

*  "  e1-165  +  e-1-165  -  2.48     ~  "  1.038"      °-UU°; 

.*.     z  =  z2-k  =  1.165  -  0.005  -  1.16, 

an  approximation  close  enough,  the  value  found  from  the 
table  by  interpolation  being  the  same. 
Hence, 

25        25        01  „u 

a  =  —  =  — —  =  21.oo  ft. 
z       1.16 

H  =  wa  =  20  X  21.55  =  431  lbs. 

Vl  =  Vs{2  +  a2=  V(31)2  + (21.55)-  =  37.75  ft. 

(6)     Cor.  Ill,  Art.  146. 

d  =  yl-  a  =  37.75  -  21.55  =  16.20  ft, 

T  =  wyt  =  20  X  37.75  =  755  lbs.     Cor.  II,  Art.  146. 

Example  2.  —  To  get  the  correction  for  the  sag  in  measur- 
ing horizontal  distances  with  a  steel  tape  unsupported  except 
at  the  ends  where  pull  is  applied. 

A  100  ft.  steel  tape  was  weighed  and  its  weight  found  to 
be  If  lbs.,  making  w  =  0.0175  lb. 

Using  the  equation  of  the  parabola, 

wx~ 
y  =  x-p,  where  P  =  H  is  the  pull  in  pounds; 

d  =  ^-5,  exact  if  I  is  span,  approx.  if  I  is  arc  or  tape. 

o  1 

Length  of  tape  =  s  =  /  +  -=-= =-=-  +  •  •  • 

O  I  O  I 

?/,273  7/,475 

,  _Sd2  _  iv2ls  {  approx.  correction  for  one  I  ,. 
X  "  S  ~  ~3l  "  2TP  (tape  length,  (10),  Art.  144  J '  () 
True  horizontal  distance 

=  l  =  s  —  KTpzi  approx.,  s3  being  put  for?3.   (11),  Art.  144.     (2) 


254  INTEGRAL  CALCULUS 

In  (1)  when  s  and  P  are  given  to  find  I  a  cubic  equation 
results,  the  solving  of  which  may  be  avoided  by  putting  s3 
for  Z3,  since  they  are  nearly  equal,  the  correction  being  small. 
Substituting  the  pull  P  in  lbs.  in  (1)  or  (2),  the  correction  or 
the  horizontal  distance,  respectively,  for  each  tape  length 
will  be  found.  The  pull  may  be  measured  with  a  spring 
balance.  Since  sag  shortens  the  distance  between  tape  ends 
and  pull  lengthens  it  by  stretching  the  steel,  there  is  some 
pull  at  which  the  two  effects  are  balanced.  The  stretch  is 
given  by 

where  P  is  pull  in  lbs.,  I  is  length  of  tape,  A  is  area  of  the 
cross  section,  and  E  is  modulus  of  elasticity,  usually  expressed 
in  lbs.  per  sq.  in.;  for  steel,  E  =  30,000,000  lbs.  per  sq.  in. 

To  find  the  pull  that  will  just  balance  the  effect  of  sag, 
put  the  values  of  x  from  (1)  and  y  from  (3)  equal; 


w2V        PI  .  . 

24P~2  =  AE'  aPProximatelyi 


whence, 


P  =  \/ — sr — )  approximately. 


In  this  example  the  width  of  the  tape  was  fV  in.  and  the 
thickness  FV  in.,  making  area  A  =  0.00525  sq.  in. 


V    2 


!5(100)330  =  271bs. 


24  (1000) 

This  pull  to  balance  effect  of  sag  can  be  determined  experi- 
mentally by  marking  on  a  horizontal  surface  two  points  100 
feet  apart,  and  then  noting  the  pull  on  a  spring  balance  when 
the  ends  of  the  suspended  tape  are  exactly  over  the  points 
marked. 

150.   The  Tractrix.  —  The  characteristic  property  of  the 
tractrix  is  that  the  length  of  its  tangent  at  any  point  is 


THE  TRACTRIX  255 

constant.     Denote  the  constant  length  of  the  tangent  PT 
by  a. 

If  a  string  of  length  a  has  a  weight  attached  at  one  end 
while  the  other  end  moves  along  OX  in  a  rough  horizontal 
plane  XOY,  the  point  P  of  the  weight,  as  it  is  drawn  over 
the  plane,  will  trace  the  tractrix  APPi  .... 


Let  AO  be  the  initial  position  of  the  string  and  PT  any 
intermediate  position.  Since  at  every  instant  the  force 
exerted  on  the  weight  at  P  is  in  the  direction  of  the  string 
PT,  the  motion  of  the  point  P  must  be  in  the  same  direction; 
that  is,  the  direction  of  the  tractrix  at  P  is  the  same  as  that 
of  the  line  PT,  which  is,  therefore,  a  tangent  to  the  curve. 

To  find  the  length  and  equation  of  the  curve:  let 

PD  =  ds;    then     -PN  =  dy    and    ND  =  dx; 

ds       _  PD       _  a  m 

•'*      dy~  ~PN~       y  U 

Hence,  if  sis  reckoned  from  A  (0,  a),  then  the  length 

s=-a  pft-afcgi  (2) 

Ja   y  y 

Also,  from  the  figure, 

dy  _  y  nr     dx  _       Va2  -  y2 .  ,. 

-j-  — /  -i     or     j—  — ,  {o) 

dx  Va2  -  y2  dy  y 


256  INTEGRAL  CALCULUS 

/.  x  =  fV  -  Va2-y2^  =   -  Va2  -  y2 
J  a  y 

+  a  log  I  — — fl2  ~  y  j  (See  Ex.  20,  Exercise  XXV.) 

is  the  equation  of  the  tractrix. 

Example  1.  —  To  find  the  area  bounded  by  the  tractrix  in 
the  first  quadrant,  the  z-axis,  and  the  ?/-axis. 

A  =    f°°ydx=  -    f°Va2-y2dy     (from  (3)) 

[y  -.  /—* — 5  .  °>2  •   i  vla    m2 

=  [2Va-y+2Sm     airir- 
Example  2.  —  To  find  the  area  bounded  by  the  catenary, 
the  x-axis,  the  i/-axis,  and  any  ordinate  y. 

A=%  PV  +  e~  V  dx  =  %  \e«  -  e~»)  -  a2  sinh  -      (1) 
Z  Jo  £  a 

—  a\-\ea  —  e   a)  \  =  as, 

where  s  =  length  of  arc.     (See  (3),  Art.  146.) 
For  area  up  to  x  =  a, 

..ft! -.-:)];- f(.-I)    ■ 

Hence,  by  (1), 


A=    ('a 

Jo 


cosh  -dx  =  a2  sinh  -  ■ 
a  a 


151.  Evolute  of  the  Tractrix.  —  It  has  been  shown  in 
Art.  146,  Corollary  III,  that  the  involute  of  the  catenary  is  the 
tractrix.  Conversely,  it  may  be  shown  that  the  evolute  of 
the  tractrix  is  the  catenary.  Let  (a,  0)  be  the  coordinates  of 
C,  the  point  of  intersection  of  the  normal  at  any  point 
P  (x,  y)  of  the  tractrix  and  the  perpendicular  to  .r-axis  at  T, 
end  of  tangent,  PT  =  a.     From  the  figure  (Art.  150), 

OT  =  a  =  x  +  Va2  -  y2,  (1) 


(3) 


EVOLUTE  OF  THE   TRACTRIX  257 

and  (TC  and  PC  being  drawn) 

m^       n      a2  (3      a  f  from  similar  triangles  1         /e%. 

TC  =  V  =  -,  since  -  =  -[     PTC  &nd  PTM 

Equation  of  tractrix, 

/-s s   i      i      [~a  +  Va2  —  w2l 

x  =  -  va2  -  ?/2  +  alog   — —  • 

Eliminating  x  and  y  from  these  three  equations  gives 
a  [(3  +  Va2  -  a2] 

.    ^n  +  VFJ,  (4) 

a 
Solving  (4)  for  /3  gives  for  the  relation  between  a  and  j3, 

which  is  the  equation  of  the  catenary  with  origin  at  0;  and 
its  lowest  point  at  A,  one  end  of  the  tractrix. 

The  normal  PC  to  the  tractrix,  the  involute,  is  the  tangent 
to  the  catenary,  the  evolute,  and  is  equal  in  length  to  the 
arc  AC  (not  drawn)  of  the  catenary. 

If  the  equations  for  coordinates,  a  and  (3,  of  the  center  of 
curvature,  given  in  Art.  94,  be  used,  the  values  found  in 
(1)  and  (2)  will  result. 

Note.  —  It  may  be  shown  independently  of  Art.  146, 
Corollary  III,  that  the  catenary  is  the  curve  which  has  the 
property  that  the  line  drawn  from  the  foot  of  any  ordinate 
of  the  curve  perpendicular  to  the  corresponding  tangent  is 
of  constant  length  a. 

Thus,  let  6  be  the  angle  which  the  tangent  CP  makes  with  the 
a:-axis  and  it  is  evident  from  the  figure  (CP  being  drawn)  that 
o.l  1 

=  COS0  = 


0  Vl+tan20 


vMS 


258  INTEGRAL  CALCULUS 

da  d/3 


l-f"da=r  d& 

CI  t/0  J  a 


V(32-a2' 

,     «  =  log(/3  +  V^^)T=logf-±^Z); 
a  ja  \  a  / 

...  ei  =  g+^^,  (4) 

a 
Hence,         /3  =  =  \ea  +  e   V ,  as  before. 


CHAPTER  IV. 

INTEGRATION   AS  THE  LIMIT   OF  A  SUM.     SUR- 
FACES  AND   VOLUMES. 

152.  Limit  of  a  Sum.  —  A  definite  integral  has  been 
defined  (Art.  12|)  as  an  increment  of  an  indefinite  integral. 

It  will  now  be  shown  that  a  definite  integral  equals  the  limit 
of  the  sum  of  an  infinite  number  of  infinitesimal  increments  or 
differentials. 

Many  problems  in  pure  and  applied  mathematics  can  be 
brought  under  the  following  form : 

Given  a  continuous  function,  y  =  f  (x),  from  x  =  a  to 
x  =  b.  Divide  the  interval  from  x  =  a  to  x  =  b  into  n  equal 
parts,  of  length  Ax  =  (b  —  a)/n.  Let  xh  Xz,  Xz,  .  .  .  xn  be 
values  of  x,  one  in  each  interval;-  take  the  value  of  the  function 
at  each  of  these  points,  and  multiply  by  Ax;  then  form  the  sum: 
f(x,)Ax+f(x2)Ax  +  •  •  •  +f(xn)Ax.  (1) 

Required,  the  limit  of  this  sum,  as  n  increases  indefinitely  and 
Ax  approaches  zero.' 

This  problem  may  be  interpreted  geometrically  as  the 
problem  of  finding  the  area  under  the  curve  y  =  /  (x), 
between  the  ordinates  x  =  a  and  x  =  b;  each  term  of  the 
sum  representing  the  area  of  a  rectangle  whose  base  is  Ax 
and  whose  altitude  is  the  height  of  the  curve  at  one  of  the 
points  selected. 

Let  the  area  M1P1BX  be  denoted  by  A;  let  OMx  =  a, 
OX  =  b,  and  PiB  be  the  locus  of  y  =  f  (x). 

Let  Az  be  one  of  the  equal  parts  of  MiX,  although  the 
parts  need  not  be  made  equal  provided  the  largest  of  them 
approaches  zero  when  n  is  made  to  increase  indefinitely. 

259 


260 


INTEGRAL  CALCULUS 


It  is  easily  seen  that  the  difference  between  the  sum  of  the 
rectangles  as  formed  and  the  area  A  is  less  than  a  rectangle 
whose  base  is  Ax  and  whose  altitude  is  a  constant,  /  (6)  — 
/  (a) .  Since  this  difference  approaches  zero  as  Ax  =  0.  the 
sum  of  either  set  of  rectangles  approaches  the  area  A  as  a 


0  My    Mz   M5  M4   M5  Mn  X 

limit.     It  is  evident  that  the  sum  of  the  rectangles  which 
are  partly  above  the  curve  is  greater  than  A,  while  the  sum 
of  those  which  are  wholly  under  the  curve  is  less  than  A. 
By  the  notation  of  a  sum,  letting  T  be  the  difference, 

^  =  2  f(x)&x±T,  where  T,  < /(b)  Ax,   =  0,  asAx  =  0; 

a 

:.      limit  £*  /  (x)  Ax  -  A  =    f  *  /  (x)  dx.  (2) 

The  equation  (2)  is  true,  for  it  has  been  already  shown  that 
A  =  fbf(x)dx=  ff(x)dx]     -  ff(x)dx\     =F(6)-F(o), 

da  *J  Jx=b     *J  Jr=a 

where  f  f(x)dx  =  F  (x).  It  follows  that  the  limit  of  (1)  is 
limr/(x,)Ax+/(x2)Ax+-  •  •  +/(xn)Ax]  =  F(6)-F(a),  (3) 
where  a  and  b  are  end  values  of  x  and   I  f(x)dx  =  F(x). 


THE  SUMMATION  PROCESS 


261 


X  >4X<- 


iraz 


The  theorem  of  this  Article  summarized  in  equation  (2)  may- 
be said  to  be  the  fundamental  theorem 
of  the  Integral  Calculus. 

As  a  simple  example  of  the  determi- 
nation of  an  area  by  getting  the  limit 
of  a  sum  of  an  indefinite  number  of  in- 
finitesimal elements  of  area,  let  a  circle 
of  radius  a  be  divided  into  concentric 
rings  of  width  Ax;  then  for  the  area 
A, 

A  =  limit  V  2ttx&x  =  2ir  (axdx  =  2irx2/2\  = 

Ax=0     ^T0  Jo  JO 

Here        A  A  =  2  w  (x  +  \  Ax)  Ax  and  dA  =  2  irx  dx. 

153.  The  Summation  Process.  —  On  account  of  the 
frequency  of  the  occurrence  of  the  summation  process,  it 
may  be  said  that  an  integral  means  the  limit  of  a  sum,  the 
limit  being  in  most  cases  most  easily  found  as  an  anti- 
differential  or  anti-derivative;  that  is,  by  the  inverse  process 
to  differentiation,  namely,  by  integration. 

The  symbol    /  for  integration,  the  elongated  S,  is  derived 

from  the  initial  letter  of  summa,  the  integral  being  originally 
conceived  as  a  definite  integral,  the  limit  of  a  sum.  Accord- 
ing to  Art.  128,  the  indefinite  integral  also  may  be  regarded 
as  the  limit  of  a  sum. 

The  fact  that  the  summation  of  an  indefinitely  large 
number  of  indefinitely  small  terms  is  in  most  cases  easily 
effected  by  a  comparatively  simple  process  is  of  the  highest 
importance.  Thus  integration  replaces  the  tedious  and  often 
difficult  process  of  direct  summation  and  gives  an  exact 
result,  while  the  other  often  gives  but  an  approximation  at 
the  best. 

While  the  process  of  summation  has  been  illustrated 
geometrically  by  the  determination  of  an  area,  the  reason 


262  INTEGRAL  CALCULUS 

of  the  process  by  no  means  depends  upon  geometrical  con- 
siderations. The  method  is  applicable  to  the  determination 
of  the  limit  of  the  sum  of  small  magnitudes  of  all  kinds  — 
volumes,  masses,  velocities,  pressures,  heat,  work,  etc.  For 
an  example  of  finding  the  limit  of  the  sum  of  small  volumes, 
consider  the  volume  V  generated  by  revolving  the  area' 
M1P1BX  of  the  figure  of  Art.  152  about  OX  as  an  axis. 
Each  of  the  rectangles  PiM2,  .  .  .  ,  PnX  will  generate  a 
cylinder  whose  volume  will  be  expressed  by  w  (/  (x))2  Ax; 
hence, 

V  =^\{f(x))2Ax  +  T, 

a 

where  T,   <  w  (/  (6))2  Ax,    =0,   as  Ax  =  0; 

/.     limit  V67T  (/  Or))2  Ax  =  V  =    fV  (f(x))2  dx.     (1) 

Ax=0  a  *J  a 

Example.  —  Find  volume  of  a  sphere  by  revolution  of 


=  a2  —  x2  :  V  =    I    7r  (a2  —  x2)  dx  =  f  to?. 


For  another  example  of  finding  the  limit  of  the  sum  of 
small  volumes,  find  the  volume  of  the  sphere  considered  as 
made  up  of  concentric  shells  of  thickness  Ap. 

V  =  lim  y.^Trp2-AP  =  4tt  f  f?dp=Sira*. 

Ap=0  «/0 

154.  Approximate  and  Exact  Summations.  —  When  the 
rate  of  change  (or  the  derivative)  of  a  variable  quantity  is 
given,  the  total  amount  (or  the  integral  of  the  rate)  can  be 
obtained  approximately  by  direct  summation,  and  exactly 
by  finding  the  limit  of  a  sum;  that  is,  by  integration. 

For  example,  suppose  the  speed  of  a  train  is  increasing 
uniformly  from  zero  to  60  miles  per  hour,  in  88  seconds; 
that  is,  from  zero  to  88  ft.  per  sec.  in  88  seconds,  the  increase 
in  speed  each  second  (the  acceleration)  is  1  foot  per  second. 


APPROXIMATE  AND  EXACT  SUMMATIONS        263 

Hence  the  speeds  at  the  beginnings  of  each  of  the  seconds 
are  0,  1,  2,  3,  ...  ,  etc. 

Taking  the  speeds  as  approximately  the  same  during  each 
second  as  at  the  beginnings,  the  total  distance. 

07.00 
s=0+l+2+3+  ■  •  •  +  86 +87  =  ^^=3828  ft., 

which  is  evidently  less  than  the  true  distance. 

Taking  the  speed  at  the  end  of  a  second  as  that  during 
the  second, 

s=l+2  +  3+4+--.  +  87  +  88  =  E^H  =  3916  ft., 

which  is  evidently  greater  than  the  true  distance.  These 
values  for  the  distance  differ  by  88  ft.  and  it  is  certain  that 
the  true  distance  is  between  3828  ft.  and  3916  ft.  When  the 
length  of  the  interval  during  which  the  speed  is  taken  as 
constant  is  reduced  more  and  more,  the  result  will  be  more 
and  more  accurate,  nearer  and  nearer  to  the  true  distance. 
Manifestly,  the  exact  distance  is  the  limit  approached  by 
this  summation  of  small  distances  as  the  interval  of  time 
At  approaches  zero: 

1=88  l=3  /»f=S8  /2~|*=88 

=  limYf>A*=    /        tdt  =  -\       =  3872ft. 
=0  M=0  ^  Jt=0  4jt=0 

In  general, 

s  =  limit  X  v\t=    f     atdt  =  §  at2, 

At=0      t=0  J  1  =  0 

a  being  constant  acceleration. 

In  mechanics,  the  determinations  of  centers  of  gravity, 
centers  of  pressure,  moments  of  inertia,  varying  stress,  etc., 
involve  the  summation  principle;  and  the  greater  number 
of  the  integrations  in  practice  appear  more  naturally  as 
limits  of  sums  than  as  reversed  rates,  anti-derivatives,  or 
anti-differe  ntials. 

The  summation  of  an  infinite  number  of  terms  is  always 


264  INTEGRAL  CALCULUS 

involved  when  one  of  the  factors  entering  into  the  problem 
varies  continuously.  For  example,  in  the  problem  of  finding 
the  mass  of  a  body,  defined  as  the  product  of  density  and 
volume;  when  the  density  p  varies  continuously, 


m 


limit  VPA7=    fpdV, 

A7=0  J 


where  the  integral  taken  between  "  limits,"  that  is,  with 
end  values  for  the  independent  variable,  is  the  limit  required. 
Hence  the  mass  is  given  by  a  definite  integral,  which  can  be 
evaluated  when  the  density  p  is  a  known  function  of  the 
volume  V,  that  is,  of  the  variables  x,  y,  z  or  r,  6,  <f>,  in  terms 
of  which  the  volume  may  be  expressed.  When  the  density 
p  is  constant,  it  is  evident  that  the  mass  is 


m 


=  5)PA7  =  p  CdV  =  PV. 


Thus,  when  the  body  is  composed  of  different  liquids  of 
varying  densities  in  the  layers  or  strata,  the  total  mass  is 
found  by  the  addition  of  a  finite  number  of  terms.  For  if 
Vi,  V2,  73,  .  .  .  Vn  denote  the  volumes  of  the  separate 
parts,  and  pi,  p2,  p3,  •  .  •  pn  the  corresponding  densities,  then 

m   =   piVl  +  p2V2  +  P3V3  +     •    •    •     +  PnVn, 

where  the  summation  is  made  without  integration. 

The  above  will  give  an  approximate  result  even  when  the 
density  varies  throughout  the  whole  mass.  When,  however, 
the  density  varies  continuously  as  in  the  atmosphere,  the 
total  volume  is  divided  into  n  parts  each  equal  to  AV  and 
each  part  is  multiplied  by  the  density  at  that  part  of  the 
body.  There  are  then  n  elements  of  the  form  pAF,  and 
when  n  is  finite  their  summation  will  be  an  approximation 
to  the  mass  of  the  whole;  but  to  get  the  exact  value,  the 
limit  of  the  sum,  as  n  becomes  infinite  and  A.V  =  0,  must  be 
found,  and  hence  the  exact  value  of  the  whole  mass  is  deter- 
mined by  the  process  of  integration. 


7 


APPROXIMATE  AND  EXACT  SUMMATIONS        265 
Example  1.  —  If  y  =  x2,  find  V  x2  Ax  for  different  values 

2  2 

of  Az,  and  get  lim  V  x2  Ax.     Get  lim  V  x2  Ax. 

A^-^n   "i  A~-^n  ^"^a 


Ax=0       '1 

When  Az  =  0.2, 


Ax=0 


]£"x2  Az  =  (l2  +  F22  +  O2  +  ]L62  +  D?)  0.2  =  2.04. 

When  Az  =  0.1, 

2)2  x2  Az  =  (l2 "+  LI2  +VL22  +  •  •  •  +  L92)  0.1  =  2.18. 

When  Ax  =  0.05, 

V%2  Az  =  (l2  +  L052  +  Li2  +  •  •  ■  +  L952)  0.05  =  2.26. 
**\ 

Lim  T  a;2Aa;=    /   x2  dx  =  -=■     =  — - — 

=  2.33J  square  units  in  MiP1P2M2. 

x-^1  Cl  xz~\l      1 

Lim  2j  x2  Ax  =    I    x2  dx  =  —     =  - 

^o  Jo  ojo       o 

=  0.33 J  =  J  of  rectangle  OilfiP^i. 


1  x-n4  Ax 

Example  2.  —  If  y  =  -,  find  2  —  f°r  different  values  of 

x  ^^  i  x 


^Ax 


lAx 


Ax.  and  get  lim  V  —     Get  V  - —  as  Ax  =  0 


266  INTEGRAL  CALCULUS 

When  Ax  =  1, 

2rNi+^)(1)  =  1-833- 

X4  At 
—  =  1.593. 
i  x 

When  Ax  =  0.1,  V  —  =  1.426. 

Lim  T  —  =    /    —  =  logrc     =  log4  -  log  1  =  log  4 
=  1.386  =  Area  AfiPJFW*. 
LimT  —  =  /   —  =  log#     =logl-loga=  -loga        =  oo  ; 

Ax=0        a    x         *Ja     x  Ja  Ja=0 

hence,  when  a  =  0,  the  limit  does  not  exist,  asY  —         =  oo . 

'  **0   X  Jax=0 

(Compare  Ex.  11,  Art.  135.) 
Note.  —  For  examples  of  application  see  Art.  189. 

EXERCISE   XXX. 

1.   If  y  =  x,  find  ^  x  Ax>  when  Ax  =  1;    when  Ax  =  0.5;   when 


'3 
,7 


Ax  =  0.2.    Get  lim  V   x  Ax.  Ans.  18;  19;  19.6. 

Ax=0  *"*3 

Arcs.  20. 


2.    If  y  =  tan  0,  find   ^    tan  0  A0,  when  Ad  =  ^ ;  when  A0  =  ^ ; 


when  Ad  =  r^.     Get  lim  V3  tan  0  Afl.     Ans.'  0.316;  0.328;  0.340. 
180  A0i°^  An*,  log.  V2  =  0.346. 

Determine  the  following  quantities  (a)  approximately  by  summation 
of  a  limited  number  of  terms;  (b)  exactly  by  finding  the  limit  of  the  sum 
of  an  infinite  number  of  terms  by  integration. 

3.  The  area  under  the  curve  y  =  x3,  from  x  =  0  to  x  =  2;    from 

X  m  —  1  tO  X  ■■  1. 

4.  The  distance  passed  over  by  a  body  falling  with  constant  accelera- 
tion g  =  32.2  per  sec.2,  from  /  =  1  to  /  =  4,  v  =  gt  being  the  relation 
of  v  and  t. 


VOLUMES 


267 


5.  The  increase  in  speed  of  a  body  falling  with  acceleration  of 
g  =  32.2  per  sec.2,  from  t  =  0 tot  =  3. 

6.  The  number  of  revolutions  made  in  5  minutes  by  a  wheel  which 
revolves  with  angular  speed  u  =  t-/ 1000  radians  per  second. 

7.  The  time  required  by  the  wheel  of  Ex.  6  to  make  the  first  ten 
revolutions. 

155.  Volumes.  —  The  volumes  of  most  solids  may  be 
found  approximately  by  the  summation  of  a  finite  number  of 
parts  and  exactly  by  finding  the  limit  of  the  sum  of  an  infinite 
number  of  terms  by  integration. 


Example.  —  To  find  the  volume  of  the  right  circular  cone 
whose  altitude  is  h  and  the  radiiis  of  whose  base  is  a.  Divid- 
ing the  volume  into  parts,  each  AT',  by  passing  planes  Ax 
apart  parallel  to  the  base  Ah,  and  denoting  a  section  at  a 
distance  x  from  the  vertex  at  the  origin  by  Ax,  then,  since 
Ax  'Ai_  =  x2  7?.'2,  V  is  given  approximately  by 


and  exactly  by 


.    x' 

AkJr2\x 


V  =  lim 

=  Ak  r* 
h1   3 


*         x 
o        h- 


"   fe2  Jo 


x2dx 


(i) 

(2) 
(3) 


While  AT  is  a  frustum  of  the  cone,  dV  may  be  represented 
by  the  cylinder  PMMi  =  Ax  •  Ax  =  iry2dx. 

It  is  to  be  noted  that  the  equations  all  apply  to  a  pyramid 
with  any  plane  base  Ah  as  well  as  to  the  cone. 


268 


INTEGRAL  CALCULUS 


For  another  example :  to  find  the  volume  of  a  sphere  with 
radius  a,  divide  by  planes  perpendicular  to  OX;  then,  since 


V  =  lim  X^o^—^  Az  =  ^  fa  (a2  -  z2)  dx 

Ax=0      -a  &  a    J-a 

=  Ma2x  ~  f L  =  $  •  h3  =  ra3> where  4» 


ira1 


Otherwise; 

2}  AF  =  2)  Ax  Aa:  =  X  ^  A-T>  where  Ax  =  iry2; 

:.     V  =  lim  V  7T?/2  Ax  =  7r  /     (a2  —  x2)  dx  =  « 7ra3. 

Ax=0  ^  J-a  u 

156.  Representation  of  a  Volume  by  an  Area.  —  In  Art. 
138  on  the  significance  of  an  area  as  an  integral  it  was  stated 
that  the  integrals  represented  by  areas  might  be  functions 
of  various  kinds.  To  show  an  example  of  a  volume  as  an 
integral  represented  by  an  area  under  a  curve,  let  the  volume 
of  the  paraboloid  of  revolution,  between  x  =  0  and  x  =  4, 
be  first  found  as  the  limit  of  the  sum  of  the  parts  between 


REPRESENTATION  OF  A  VOLUME  BY  AN  AREA   269 

the  parallel  planes  Ax  apart,  as  Ax  =  0  and  the  number 
of  the  parts  increases  without  limit.  The  equation  of  the 
generating  parabola  being  y2  =  J  x, 

V  =  limit  V  wy2  Ax  =  ■?  I    x  dx  =  £  tt     =  2tt  cubic  units. 
ax=o  fie,  4  Jo  4  2  Jo 


*-X 


To  represent  this  volume  graphically  by  an  area,  the  line  OP' 

is  drawn  by  the  equation  y  =  jx,  this  being  the  function 

which  was  integrated  to  get  the  volume  of  the  solid  P2OP'2. 

Producing  the  ordinate  M2P2  to  P",  the  area  OM2P" 

graphically  represents  the  volume  of  the  solid  P2OP'2.     For, 

ydx=-r  I    xdx=-i-Fr\  =  2  7T  square  units. 

0  4  Jq  4    Zjo 


270  INTEGRAL  CALCULUS 

The  last  result  may  be  verified  by  noting  that  the  ordinate 
M%P" ,  for  x  =  4,  being  w,  the  area  of  the  triangle  is  2  t. 

In  the  same  way  it  may  be  seen  that  any  part  of  the  area, 
as  OMP',  represents  the  corresponding  part  of  the  volume 
of  the  solid ;  that  is,  there  is  the  same  number  of  square  units 
in  the  one  as  there  are  cubic  units  in  the  other. 

If  OP'P'",  the  first  integral  curve  of  OP'P",  whose  equa- 
tion is 

y  =    /    ^rxdx  =  -=-  (see  Art.  140) 

Jo    4  o 

be  drawn,  its  ordinates  will  represent  both  the  areas  of  the 
parts  of  OM2P"  and  the  volumes  of  the  parts  of  the  parab- 
oloid measured  from  0;  that  is,  the  measure  of  the  ordinates 
in  linear  units  will  be  the  same  as  that  of  the  areas  in  square 
units  and  that  of  the  volumes  in  cubic  units. 

Length  of  M2P'"  =  y  =  -«-        =  2tt  linear  units. 
Note.  —  The  volume  of  the  cone  of  Art.  155  may  be  graphi- 
cally represented  by  the  area  under  the  parabola  y  =  -jjx2, 

and  the  volume  of  the  sphere  by  the  area  under  the  parabola 
y  =  7T  (a2  —  x2).     If  the  first  integral  curves, 

V  =  g-p a?    and    y  =  v \o~x  -  |- j , 

be  drawn,  their  ordinates  will  represent  both  the  areas  and  the 
volumes  in  the  two  cases,  respectively. 

157.  Surface  and  Volume  of  Any  Frustum.  —  A  solid 
bounded  by  two  parallel  planes  is,  in  general,  called  a 
frustum.  One  or  both  of  the  truncating  planes  may  in 
special  cases,  as  in  the  sphere,  touch  the  frustum  \n  only 
one  point  and  be  tangent  planes. 

The  method  of  dividing  the  solid  into  thin  slices  and  taking 
the  sum  of  the  approximate  expressions  for  the  small  parts 
as  an  approximate  expression  for  the  whole,  and  taking  the 


SURFACE  AND  VOLUME  OF  ANY  FRUSTUM       271 


limit  of  the  sum  as  an  exact  expression  for  the  whole,  may  be 
applied  to  any  solid  even  when  the  solid  is  not  regular  and 
the  sections  not  regular  plane  curves. 


Let  the  solid  represented  in  the  figure  be  divided  into 
slices  by  planes  perpendicular  to  an  axis  OX;  then,  taking 
Ax\x  as  an  approximate  expression  for  the  volume  of  the 
slice  P  —  N1M1R1,  Ax  being  the  area  of  the  section  PNMR 
at  a  distance  x  from  plane  ZOY,  the  approximate  expression 
is 

x=h 


2)A7  =  2)^*^i 


x=0 


where  h(=  OA)  is  the  distance  between  the  truncating  or 
bounding  planes.     The  exact  expression  is 

V  =  Urn  V.  AxAx  =    /       Axdx.  (1) 

Ax=0  £*Q  Jx=Q 

When  the  area  of  a  section  is  a  function  of  the  distance  x 
from  one  of  the  bounding  planes  and  hence  Ax  can  be  ex- 
pressed in  terms  of  x,  the  limit  may  be  found  by  integration. 
The  frustum  formula  for  volumes  is,  therefore, 


-x 


x=h 


F  (x)  dx    or     V 


Jx  =  hx 


F{x)dx,        (!') 


272  INTEGRAL  CALCULUS 

where  Ax  is  F(x),  some  function  of  x;  the  one  form  giving 
the  whole  volume  and  the  other  a  segment  or  any  part 
thereof. 

To  get  expressions  for  the  area  of  the  surface  S,  let  P  be 
the  curve  NPR,  then  A£  =  NPRRiP'Nh  and  the  approxi- 
mate expression  is 

X^S=  £PAs, 

s=0 

where  As  =  NNi  and  s  is  the  length  of  CN. 
The  exact  expression  for  the  surface  is 

£  =  limit  £P  As  =    /        Pds.  (2) 

As=0   s=0  Js=0 

When  the  curve  P  is  a  function  of  s,  the  bounding  curve  in 
XZ  plane,  and  can  be  expressed  in  terms  of  s,  or  when  ds 
can  be  expressed  in  terms  of  P,  with  change  of  end  values, 
the  limit  can  be  found  by  integration.  If  the  surface  S  is 
conceived  as  generated  by  the  curve  NPR  as  it  moves  with 
its  plane  always  perpendicular  to  OX,  when  its  plane  is  in 
the  position  as  shown,  at  a  distance  x  from  plane  YZ,  let 
iViV'i  be  drawn  equal  to  ds  but  parallel  to  OX;  then  since  the 
surface  is  cylindrical,  the  increase  of  S,  if  the  increase  became 
uniform,  is 

dS  =  P  ds,    the  surface  NPRR'P'Ni'; 


8 


Jr*s=sh 
Pds.  (2) 

s=0 

If  the  curve  NPR  is  a  circle,  as  in  solids  of  revolution,  with 
the  center  at  M  on  the  z-axis,  then  P  =  2  iry  and  Ax  =  iry2, 
(2)  and  (1)  becoming 

yds  (3) 

=0 

Jn*~h 
y2  dx,  (4) 

x=0 


SURFACE  AND  VOLUME  OF  ANY  FRUSTUM   273 

where  dV  =  wy2  dx  is  the  volume  of  the  cylinder  generated 
by  the  area  of  the  circle  wy2,  as  it  moves  uniformly  through 
Az  =  dx. 

Note.  —  In  deriving  (1)  and  (4),  in  the  figure,  NNi  =  Az  = 
dx;  while  in  deriving  (2)  and  (3),  ds,  the  uniform  change  of 
s  along  a  tangent  to  the  curve  CN  at  the  point  A7,  is  drawn 
parallel  to  OX  and  represented  in  length  by  NNi,  although 
it  is  not  the  same  as  Az  =  dx  but  is  really  longer. 

Example  1.  —  To  find  the  lateral  surface  of  the  cone  of 
Art.  155:  by  (3), 

S  =  2 7r  /     yds  =  27rj        -y  sds,  where  s  =  -y  =  OP, 

Jo  I/O  '  a 

as2~\l 
-=  2Trj7r\   =  wal,  where  I  =  OPh,  an  element. 
I  Z  Jo 

Again, 

y  ds  =  2  7r   I        y-dy,  since  c?s  =  d  ( -  y )  =  -  dy, 

o  Jo  fl  \a  /     a 

limit  or  end  value  being  changed  from  2  to  a. 

Example  2.  —  To  find  the  surface  of  the  paraboloid  of 
Art,  156: 


=  2 t £ y  [1  +  64  jf ]%     from  y2  =  \x, 


dy 


2 

128 


|[l  +  (64^]J  =^((65)1-1) 


o^  (65  V65  —  l)  square  units. 


274 


INTEGRAL  CALCULUS 


Example  3.  —  To  find  the  surface  of  the  sphere  of  Art.  155, 
or  any  part  of  it,  as  a  zone. 

For  a  change  take  origin  at  A'  on  the  circumference, 
making  y  =  V2  ax  —  x2  and  2  iry  the  curve  P  bounding  the 
section  Ax;  then  by  (2)  or  (3), 

S=    \Pds  =  2ir  j      yds  =  2tt  \    adx 

JO  Jo  Jx0 

(where  yds  =  adx,  from  similar  triangles,  OMP  and  PDT) 

]x  ~|2a 

=  2  7ra  (x  —  Xo)    or    2  wax      =  4  ira2. 
x0  JO 


Ddx  T 


D'      T' 


Drawing  PT  =  ds  from  P  parallel  to  rr-axis,  2  wy  ds  is  the 
lateral  surface  of  the  cylinder  PT',  which  is  equal  in  area 
to  that  of  the  cylinder  DT' ',  Which  is  2  ira  dx. 
The  volume  is  again,  with  origin  at  A',  by  (4),1 


V=ir  J      y2dx  =  ir  f      (2 ax  -  x2) dx  =  tt  — ^ ^       = 


7r(7,\ 


Example  4.  —  To  find  the  lateral  surface  of  a  quadrangular 
pyramid.  Let  Ph  =  perimeter  of  base  and  I  =  0Ph  =  slant 
height.  Let  PMN  be  the  position  of  the  generating  perim- 
eter P  when  s  =  OP.    Since  P  and  Ph  are  similar, 


SURFACE  AND  VOLUME  OF  ANY  FRUSTUM   275 


P        OP       s       ,  „       Ph      .     ,.. 

ph  =  0Fh  =  V    hence'   p  =  Ts'm(2); 

that  is,  the  convex  surface  of  any  pyramid  or  cone  (Ex.  1) 
is  measured  by  half  the  product  of  perimeter  of  base  and 
slant  height. 

Y 


For  the  volume, 


V 


Jo 


Axdx 


=  AuCh 
"  ft2  Jo  " 
=  Ah  ff       1 
h?    3  Jo      3 


since  -r-  = 


A, 


a? 

ft2' 


Ax  =  areaPMiV, 


aAhh; 


that  is,  the  volume  of  any  pyramid  or  cone  ((3),  Art.  155)  is 
measured  by  one-third  the  product  of  its  base  and  altitude. 

Note.  —  The  foregoing,  for  the  purpose  of  illustration,  have 
been  for  the  most  part  examples  of  elementary  solids  whose 
surfaces  and  volumes  are  known  from  solid  geometry.  The 
fruitfulness  of  the  method  is  seen  in  the  determination  of  the 
surfaces  and  volumes  of  the  frusta  of  unfamiliar  and  complex 
solids. 

The  following  are  some  examples: 

Example  5.  A  monument  is  to  be  built  in  horizontal  rec- 
tangular sections,  one  side  of  a  section  to  vary  as  the  dis- 
tance below  the  top  and  the  other  as  the  square  of  this 


276 


INTEGRAL  CALCULUS 


distance.  The  base  is  to  be  a  square  30  feet  on  a  side,  and 
the  height  of  the  monument  is  to  be  20  feet.  Find  the  vol- 
ume when  it  is  made  up  of  rectangular  blocks  with  vertical 
sides;  and  also  the  volume  when  the  sections  vary  contin- 
uously from  top  to  base. 

Taking  origin  at  top  and  z  positive  downward,  let  Az  = 
4xy,  2y  =  az,  2x  =  bz2;  2y  =  20a  =  30;  .'.  a  =  f: ;  2x  = 
400  b  =  30;  .*.  b  =  ^ ;  .'.  y  =  %z,  for  line  OB;  x  =  & z2> 
for  curve  OA.    Az  =  4xy  =  abzz  =  fa  z3, 

2=20  -i 

V  =  X  ^Az  =5.^[0  +  53  +  103  +  153-|-203] 

3  JA*=5 

=  7031i  cu.  ft. 

V  =  lim  T  —  z3Az  =  —  / 


z3dz 


9^  [VI2 
80  |_4  Jc 


4500  cu.  ft. 


Hence,  as  Az,  the  thickness 
of  the  blocks,  is  made  less 
and  less,  the  volume  will 
approach  4500  cu.  ft.,  the 
volume  when  sections  vary 
continuously.  The  plan  is 
shown  on  reduced  scale. 
The  projection  of  OD,  the 
curve  of  intersection  of  the 

JO 


plane  surface  OBD  and  the  curved  surface  OAD,  is  O'D 
on  plane  parallel  to  IF  plane.     The  equation  of  O'D  is 


SURFACE  AND  VOLUME  OF  ANY  FRUSTUM        277 

y1  =  15  x,  by  eliminating  z  from  y  =  \z  and  x  =  A  z2. 
The  plan  shows  the  corners  of  the  blocks  on  this  curve. 

Example  6.  Find  the  lateral  surface  of  the  monument  of 
Ex.  5.  When  built  of  rectangular  blocks,  the  sum  of  the 
rectangular  areas  gives  the  area  of  the  surface.  When  the 
stone  is  shaped  to  make  sections  vary  continuously,  or  when 
this  is  effected  by  using  concrete  in  shaped  forms,  find  the 
areas  of  the  surfaces  OBD  and  OAD  separately. 

i(OAD)  -  ^P|^]  -  2f  x  38.8  -  862  „.  ft. 

=  8oJo    *  L1+16j  ^SO'U    "-aLsJi 

3    8000      10R        ft 
=  64'-3-=12°Sq-ft- 

4  (OBD)  =  4  X  125  =  500  sq.  ft.     Total  surface  =  1362  sq.  ft. 

Note.  —  Since  OBD  is  a  plane  surface,  its  area  may  be 
found  by 

A=iX  dS  =  Lo    125  S'ds  "  125  '  4  =  125  Sq-  ft- 

as  above.  Here  z  =  Tf  5  s2  is  the  equation  of  curve  OD  in 
the  oblique  plane  of  sx,  for  since  OB  =  25,  y  =  f  a,  and 
2/2  =  15  x  becomes  ?9:-7  s2  =  15  z,  or  x  =  Tfy  s2. 

In  Ex.  5  above,  ?/2  =  15  x  is  given  as  equation  of  pro- 
jection of  OD  on  xy  plane,  or  any  plane  parallel  thereto. 


278 


INTEGRAL  CALCULUS 


While  OD  is  given -as  a  line  in  space  by  two  equations,  by 
rotating  axis  OY  about  OX  through  tan-1  §  %  =  cos-1  f ,  it  is 
given  by  one  equation,  s2  =  1%&  x,  in  plane  of  sx. 

Example  7.  Find  the  volume  common  to  two  right  circular 
cylinders  of  equal  radius  a,  whose  axes  intersect  at  right 
angles. 


Let  the  two  cylinders  be  x2  -\-  z2  =  a2  and  y2  +  z2  =  a2; 


then  Az  =  LMPN 


xy  =  a2 


z2,  and 
V  =  8  faA2  dz  =  8  /"(a2  -  z2)  dz  =  s\a2z  -  |T  =  |a3. 

The  total  volume  common,  being  8  times  Z  —  OACB,  is  -^  a3. 

Example  8.  A  dome  has  the  shape  of  the  figure  of  Ex.  7, 
find  the  area  of  the  curved  surface. 

The  surface  ZBC  is  equal  in  area  to  the  surface  ZAC,  and 
is  one-eighth  part  of  the  surface  of  the  dome,  which  surface 
is  the  upper  half  of  the  surface  of  the  common  volume  of 
Ex.7. 

Hence  the  surface  of  the  dome  of  eight  equal  parts  is 
given  by 


SURFACE  AND  VOLUME  OF  ANY  FRUSTUM        279 

S  =  8  ZBC  =  8  fShP  ds  =  S  fShNPds 
Jo  Jo 

-°I«h(t)T* 

=  8 a  I    -dz  =  Sa  I    dz  =  8a2. 
Jo   2/  Jo 

The  result  shows  that  each  of  the  curved  surfaces  of  the  solid 
Z  —  OACB  is  equal  in  area  to 
its  base  OACB;  the  surface  of 
the  dome  being  just  twice  that 
of  its  base. 

Note.  —  Another  determina-  tfa 
tion  of  the  area  of  ZBC  may- 
be made  by  developing  the 
curved  surface  upon  a  plane 
and  finding  the  area  as  a  plane 
area,     Thus,    developing    ZBC 


?--, J  c 

as  the  plane  area  Z'BC,  with  B  as  origin; 


area  Z'BC  =  area  ZBC 


TTU, 

f2  x'dz', 


where  x  =  x'  =  a  cos  6,  and  z'  =  ad 

IT  TT 

=    I    a  cos  Bd  {ad)  =  a2  sin  0     =  a2. 
Jo  Jo 

Example  9.  Given  a  right  cylinder 
of  altitude  h,  and  radius  of  base  a. 
Through  a  diameter  of  the  upper  base 
two   planes    are    passed,    touching   the 

lower  base  on  opposite  sides.     Find  the  volume  included 

between  the  planes. 


280 


INTEGRAL  CALCULUS 


V  =    r  Axdx  =  4  f°  (MNPR)  dx  =  4  Pyz  dx 

t/0  t/0  «/o 

=  4   P(a2-z2)^-(a-z)<fo 

t/0  a 

=  —   Pa  (a2  -  x>)*  dx-—  P  (a2  -  xrf  x  dx 
a  Jo  a  Jo 

= 4f[}vy=^+}*r*5|;+g^  -  *>»]; 


ira?h—  -  a2/i  =  (vol.  of  cylinder)  —  (vol.  outside  the  planes), 
o 


Here 
NP 


MR  =  OZ 


MA 
OA  ' 


NP 


z  =  -  (a  —  x) ; 
a 


MN  =  RP  =  y  =  (a2  -  x2)*. 


It  may  be  noted  that,  when 
h  is  equal  to  a,  the  volume 
outside  of  the  planes  being  % 
a3,  is  one-fourth  of  the  volume 
common  to  the  two  cylinders  of 
Ex.  7. 

Example  10.  Two  cylinders 
of  equal  altitude  h  have  a 
circle  of  radius  a,  for  their 
common  upper  base.  Their 
lower  bases  are  tangent  to  each 
other.  Find  the  volume  com- 
mon to  the  two  cylinders. 


V=    fhiAvdy=   P(PMM')dy  =    Pxzdy 

J  ht  J  —a  J— a 

=    /    x-xdy  =  -  I    x2dy  =  -   I    (a2  —  y2)  dy 

J —a      0,  d  J  —a  Q>  J— a 

=  a[ay~^L  =  3ah- 


SURFACE  AND  VOLUME  OF  ANY  FRUSTUM        281 


Here,  PMMf  being  similar  to  ZAA', 

Ar„      OZ-NM  h        , 

NP  =  — tti —  >     or    z  =  -x,  where  x2  =  a2  —  y2. 
OA  a  u 

P  is  on  curve  of  intersection  of  the  cylinders. 

It  is  seen  that  the  volume  found  is  equal  to  the  volume 
outside  the  planes  of  Ex.  9. 

Example  11.  A  torus  is  generated  by  a  circle  of  radius  b 
revolving  about  an  axis  in  its  plane,  a  being  the  distance 
of  the  center  of  the  circle  from  the  axis. 


Find  the  volume  by  means  of  sections  perpendicular  to 
the  axis. 

V  =  fkiAvdy=  fXh  [T(a  +  x)2-Tr(a-x)2]dy 

Jhx  Jx=-b 

-  7T  fU  b  [(a  +  Vb2  -  y2)2  -  (a-  Vb2  -  y2)2]  dy 
=  tt  /     4a  Vb2  -  y2dy 

0  = 


=  47ra 
=  47ra 


"Trfe2" 
2 


2w2ab2  =  2ira-Trb2. 


282  INTEGRAL  CALCULUS 

Note.  —  The  last  form  of  the  result  shows  that  the  .volume 
is  the  product  of  the  area  of  the  cross  section  and  the  length 
of  the  circumference  described  by  the  center  of  the  revolving 
circle,  radius  a  being  mean  of  a  +  b  and  a  —  b. 


EXERCISE   XXXI. 

1.  Find  the  volume  of  the  right  conoid  whose 
base  is  a  circle  of  radius  a,  and  whose  altitude 
is  h. 

(a)  With  origin  at  0,  on  the  circumference; 
y2  =  2  ax  -  x2. 

a2  -  x2. 

Aiis.   — • 

2.  An  isosceles  triangle  moves  perpendicular  to  the  plane  of  the 
ellipse  x2/a2  +  y2/b2  =  1,  its  base  is  the  double  ordinate  of  the  ellipse, 
and  the  vertical  angle  2  A  is  constant.  Find  the  volume  generated  by 
the  triangle.  .         4  ab2  cot  A 


Ans. 


3 


x2      y2      s2 

3.  Find  the  volume  of  the  ellipsoid  i  +  jb  +  "^  =  1  by  considering 

the  volume  generated  by  moving  a  variable  ellipse  along  the  axis  of  X. 
Area  of  ellipse  =  irab.     From  result  get  volume  of  a  sphere. 

Ans.  %irabc. 

4.  A  football  is  16  inches  long  and  a  plane  section  containing  a  seam 
is  an  ellipse  the  minor  axis  of  which  is  8  inches  in  length.  Find  the 
volume  (a)  if  the  leather  is  so  stiff  that  every  cross  section  is  a  square; 
(b)  if  the  cross  section  is  a  circle.  Ans.  (a)  341  §  cu.  in. 

,..      512  7T 

(6)  — ^—  cu.  in. 

6.  To  fell  a  tree  2  a  feet  in  diameter,  a  cut  is  made  halfway  through 
from  each  side.  The  lower  face  of  each  cut  is  horizontal;  the  inclined 
face  makes  an  angle  of  45°  with  the  horizontal.  Find  the  volume  of 
the  wood  cut  out.     Compare  Ex.  9  of  illustrative  examples. 

Ans.  f  a3  cu.  ft. 

y2      z2 

6.  Find  the  volume  of  the  elliptic  paraboloid  2  x  =  —  H —  cut  off 

by  the  plane  x  =  h.  Ans.  ir  "vpq  h2. 

7.  Find  the  volume  of  Ex.  9  by  moving  the  trapezoidal  section  along 
the  F-axis.     Note  that  the  triangular  section  of  the  volume  outside  the 


PRISMOID  FORMULA  283 

cutting  planes  will  at  the  same  time  generate  that  volume,  the  same  as 
the  volume  of  5  above,  when  h  =  a. 

8.  A  cap  for  a  post  is  a  solid  of  which  every  horizontal  section  is  a 
square,  and  the  corners  of  the  square  lie  in  the  surface  of  a  sphere  12 
inches  in  diameter  with  its  center  in  the  upper  face  of  the  cap.  The 
depth  of  the  cap  is  4  inches.  Find  the  volume  of  the  cap.  Compare 
Ex.  7  of  illustrative  examples. 

Ans.  490f  cu.  in. 

9.  Find  the  surface  of  the  cap  of  8,  above.  Compare  Ex.  8  of 
illustrative  examples. 

Ans.  Curved  surface  =  192  sq.  in.;  surface  of  top  =  72 sq.  in. 

10.  Show  that  the  volume  of  the  frustum  of  any  pyramid  or  cone  is 

equal  to  ^  (A0  +  Ah  +  VAoAh)  where  A0  and  Ah  are  the  bases,  and  h 
is  its  height. 

158.  Prismoid  Formula.*  —  If  two  solids  contained 
between  the  same  two  parallel  planes  have  all  their  corre- 
sponding sections  parallel  to  these  planes  equal,  that  is,  if 
the  area  A/  of  the  one  is  the  same  as  the  area  As"  of  the  other, 
then  their  total  volumes  are  equal,  since  the  two  volumes 
are  given  by  the  same  integral.  Let  the  distance  between 
the  bounding  planes  be,  in  general,  s  =  x,  or  y,  or  z. 

If  the  area  A8  is  a  section  of  a  solid  included  between  two 
parallel  planes  and  is  a  quadratic  function  of  s, 

As  =  as2  +  bs  +  c,  (1) 

where  s  is  the  distance  of  the  section  As  from  one  of  the  two 
parallel  planes,  then  the  volume  is  given  by 

(as2  -{- bs  +  c)  ds  =  \a  -  +  b  -  +  cs\ 

ah3      bh?        ,  ,  . 

=  -y  +  -7r  +  ch,         (2) 


Jo  Js=0 


where  h  is  the  distance  of  the  terminal  plane  from  the  initial 
plane  of  reference;  that  is,  the  height,  or  length,  of  the  solid, 
as  the  case  may  be. 

*  This  derivation  of  the  formula  is  substantially  that  given  in  Davis's 
Calculus. 


284  INTEGRAL  CALCULUS 

The  area  A0  =  A8 

the  area  Ah  =AS 

and  the  area  Am  =  Ae 


■i 


=  as2  +  bs  +  c 

=  0 

=  c; 

s  = 

=  as2  +  bs  +  c        =a/i2  +  ^  +  c; 

=^                                      Js=/i 

=  as2  +  6s  +  c]       =^-+  —  +  c, 

where  Am  is  the  area  of  a  section  midway  between  the  end 
sections,  A0  and  Ah. 

The  average  of  A0,  AA,  and  4  times  Am,  is 

g  (A0  +  Aa  +  4  ATO)  =  -y  +  -^  +  c; 


and  this  average  section  multiplied  by  /i  is  the  total  volume : 

,+A,+4Am        _  o^3     ta2 
6  Xn~  3  +  2 


Js=0 


This  is  the  Prismoid  Formula,  so  called  because  it  holds 
not  only  for  every  solid  whose  volume  is  given  in  elementary 
geometry  but  for  any  prismoid,  that  is,  for  a  solid  with  any  end 
sections  whatever,  with  sides  formed  by  straight  lines  joining 
points  of  one  end  section  with  points  of  the  other  end  section. 

159.  Application  of  the  Prismoid  Formula.  —  The  for- 
mula holds  even  for  many  solids  that  are  not  prismoids,  for 
example,  spheres  and  paraboloids.  It  holds  for  all  solids 
defined  by  equation  (1),  Art.  158,  and  even  for  all  cases 
where  A8  is  any  cubic  function  of  s: 

A8  =  as*  +  bs2  +  cs  +  d.  (1) 

When  /  (x)  is  a  quadratic  or  a  cubic  function  of  x;  then, 
in  general, 

JT|V  0»)  dx  =  [/(a)  +  4/(^±-6)  +/(6)]^>         (2) 

in  accordance  with  the  prismoid  formula.  The  practical 
application  of  the  formula  is  mainly  for  the  close  approxi- 
mation it  gives  to  the  volume  of  objects  in  nature;  for  any 


APPLICATION  OF  THE  PRISMOID  FORMULA       285 


elevation  or  irregularity  of  the  crust  of  the  earth  can  be 

approximated  to  quite  closely,  either  by  the  frustum  of  a 

cone,  sphere,  cylinder,  pyramid,  paraboloid,  wedge,  or  prism; 

and  as  the  formula  holds  for  these  solids  as  well  as  for  any 

combination  of  them,  it  can  be  applied  without  determining 

which  of  the  solids  actually  approximates  most  nearly  to 

the  object  whose  volume  is  desired.     While  it  is  thus  used 

to  approximate  to  the  volumes  of  irregular  solids,  it  is  to 

be  remembered  that  it  gives  the  exact  volume,  when  the  area 

of  a  section  As  is  either  a  quadratic  or  a  cubic  function  of  s, 

including  of  course  a  linear  function  as  a  special  case  of  the 

quadratic  or  cubic  function. 

Example  1.  —  In  the  case  of  the  cone  or  pyramid,  Art.  155, 

x2 
it  is  seen  that  Ax  =  Ah  u  is  &  quadratic  function  of  x,  and 

hence  V  =  ^(o  +  Ah  +  4^/i  =  ±Ahh. 

Example  2.  —  In  the  case  of  the  sphere, 

n%  —  1*2 


AX  =  A{ 


=  T  (a2  —  x2), 


hence,    V  =  J  (0  +  0  +  4  A0)  2  a  =  f  A0a 
=  Am  =  ira2. 

Example  3.  —  In  the  case 
of  the  paraboloid  of  revolu- 
tion, about  the  axis  OY,  of 
the  curve  y  =  x2; 

7=i(0  +  A,  +  4Am)/i 

Here  A8  =  iry  is  a  linear 
function  of  the  distance  y, 
for  by  (1),  Art.  158,  a  =  0, 
b  =  7r,  c  =  0;  hence  the 
formula  holds. 


|7ra3,    where   A0 


286  INTEGRAL  CALCULUS 

Example  4.  —  The  prismoid  shown  in  figure  is  composed 
of  a  prism,  a  wedge,  and  two  pyramids.  Let  A0  be  the 
smaller  end  section,  Ah  the  larger,  and  Am  the  mid  section. 

V  =  A0h  =  g  (A0  +  A0  +  4  A0),  for  prism, 
F  -  A'h  \  -  J(o  +  A;  +  4^),  for  wedge, 

V  =  A);  I  =  1(0  +  A;'  +  ii*) ,  for  pyramid. 

The  formula  is  seen  to  hold  for  the  three  forms  of  solids 
composing  the  prismoid. 


As  a  practical  case,  let  the  figure  represent  a  section  of  a 
railway  embankment  100  feet  in  length. 

A  fill  of  10  ft.  with  side  slopes  lj  to  1,  makes  A0  =  250  sq.  ft. 
A  fill  of  20  ft.  with  side  slopes  1^  to  1,  makes  Ah  =  800 

sq.  ft. 
A  fill  of  15  ft.  with  side  slopes  li  to  1,  makes  4  Am  =  1950 
sq.  ft. 
Hence   V=  H*  (250  +  800  +  1950)  =  50,000  cu.  ft. 


APPLICATION  OF  THE  PRISMOID  FORMULA       287 

Here  V  =  h/2  (A0  +  Ah)  =  H0-  (250  +  800)  =  52,500  cu.  ft., 
by  average  end  areas. 

V  =  luAm  =  100  X  487.5  =  48,750  cu.  ft.,  by  mean  area. 
It  is  seen  that  the  error  of  the  approximation  by  the  average 
end  areas  is  twice  that  by  mean  area  and  of  opposite  sign. 
Since  the  errors  vary  as  the  square  of  the  difference  in  dimen- 
sions of  the  two  end  areas,  when  the  end  areas  are  very 
different,  the  true  prismoid  formula  should  be  used,  but 
when  the  end  areas  are  alike,  or  nearly  so,  the  approximate 
formulas  may  give  results  as  nearly  exact  as  may  be  desired. 

EXERCISE   XXXII. 

1.  Get  the  volume  of  a  frustum  of  a  solid  included  between  the 
planes  s  =  0  and  s  =  h,  when  the  area  A8  of  a  parallel  cross  section  is  a 
cubic  function,  as3  +  bs2  +  cs  +  d,  of  the  distance  s  from  one  of  the 
bounding  planes;  first  by  direct  integration  using  the  frustum  formula, 
then  by  the  prismoid  formula.  Thus  prove  the  statement  at  the 
beginning  of  Art.  159. 

2.  Show  according  to  (2)  Art.  159,  as  in  the  case  of  volumes,  that  the 
area  under  any  curve  y  =  f  (x),  where  /  (x)  is  any  quadratic  or  cubic 
function  of  x,  between  x  —  a  and  x  =  b,  is 

— g—  (Va  +  ?/6  +  4  ym),  (1) 

where  ya,  yb,  ym  represent  the  values  of  y  at  x  =  a,  x  =  b,  and  x  = 

§(«  +  &). 

3.  Find,  first  by  direct  integration,  and  then  by  (1)  of  Ex.  2,  the 
areas  under  each  of  the  following  curves. 

(a)   y  =  x2,  between  x  =  0  and  x  =  2. 

(6)    y  =  x2  +  2  x  +  3,  between  x  =  1  and  x  =  5. 

4.  Show  that,  when  (1)  of  Ex.  2  is  used  to  get  area  under  the  curve 
y  =  x*  between  x  =  1  and  x  =  3,  the  error  is  about  4.2  per  cent. 

6.  Find  the  volume  made  by  revolving  the  area  between  the  curve 
y  =  x2  and  the  z-axis  about  the  z-axis,  between  x  =  0  and  x  =  2.  See 
Ex.  3,  Art.  159.  Find  first  by  (1)  of  Art.  153;  then  by  the  prismoid 
formula  show  that  the  result  by  that  formula  is  in  error  about  4.2 
per  cent. 

Note.  —  The  prismoid  formula  is  not  applicable  for  exact  results, 
when  A8  is  given  by  a  higher  function  than  a  cubic;  in  that  case,  it  and 
the  general,  formula  (2),  Art.  159,  for  /  (x),  give  approximations. 


288 


INTEGRAL  CALCULUS 


160.   Surfaces  and  Solids  of  Revolution.  —  To  get  an 

expression  for  the  area  of  a  surface  made  by  the  revolution 

of    a    curve    y  =  f  (x) 
//|  s  An  about  the  axis  OX,  let 

A  (xo,  y0)  be  a  fixed 
point  and  P  (x,  y)  a  va- 
riable point  on  the  curve 
OPQP.  Let  P0P  =  s, 
and  PP'  =  As,  and  let 
PD  and  P'R  be  drawn 
each  parallel  to  OX  and 
equal  in  length  to  As. 

Let  S  denote  the  surface  generated  by  the  revolution  of  P0P 

about  the  z-axis;    then  A$  equals  the  surface  generated  by 

PP'.     It  is  evident  that 

surface  PD  <  A£  <  surface  P'R) 

that  is,  2  iry  As  <  ^S  <  2  w  {y  +  ^y)  As; 

A*S 
dividing  by  As,    2  iry  <  -^  <  2  t  (y  +  Ay)  ; 

=  -T-  =  2tt?/,  since  A?/  =  0,  as  As  =  0; 

/.     dS  =  2  iry  ds  or  5  =  2  tt  /  S?/  ds.    (See  (3)  Art,  157.)     (1) 

Jo 

Here  dS  =  2iry  ds  may  be  represented  by  the  lateral  surface 
of  a  cylinder  MPT',  the  circumference  of  whose  base  is  2  7r?/ 
and  whose  length  is  PT' ',  drawn  parallel  to  OX  and  equal  to 
PT,  which  represents  ds  along  the  tangent  at  P.  This  is  so, 
for  this  surface  is  what  the  change  of  S  would' he,  if  at  P  the 
change  became  uniform,  ds  being  the  uniform  change  of  s 
as  x  increases  uniformly  from  that  point.  The  surface  S 
may  be  considered  as  generated  by  the  circumference  of  a 
circle  of  varying  radius  y  and  hence  the  point  P  moving  on 
the  curve  according  to  the  law  expressed  by  its  equation 
y=f(x).    Since 


hence,   limit , 

As=0    L^S 


TA/S 


SURFACES  AND  SOLIDS  OF  REVOLUTION 


289 


ds  =  (dx>  +  <y)i  -  [l  +  (|)2]"  dx   or  [(|)2  +  lj  dy, 
(1)  becomes 


— JC-L»+eB 


Wr 

<dy 


+  1 


da;, 


dy. 


(2) 
(3) 


Similarly,  when  the  ?/-axis  is  the  axis  of  revolution 

These  formulas  may  be  derived  as  the  limits  of  sums;  thus 
S  =  X  (surfaces  As)  =  lim  x   (surfaces  chord  As), 

^  As=0  ^0 


(4) 
(5) 


r. 


arc  As 


Vchord 


£** 


=  lim  yS  2  iry  As  =  lim  T*  2  ^  [~i  +  (£|Y1  Ax 

Ax=0^0  Ax=0  ^Jo  L  V**/    J 

-*-J!Ml+GJTfc  (2) 

The  other  forms  may  be  derived  in  the  same  way,  which  is 
an  abbreviation  of  a  rigorous  derivation. 

In  any  particular  example  to  which  these  formulas  are 
applicable,  use  that  form  which  involves  the  simpler  inte- 
gration. 

For  volumes  of  solids  of  revolution; 

V  =  7T  f\fdx=ir  fX(f  (x))2  dx,  (See  (4),  Art.  157)     (6) 

Jx0  Jo 

when  the  revolution  is  about  the  x-axis;  and 

y  =  7r  fUx*dy  =  ir  fV(f(y)Ydy,  (7) 

when  the  i/-axis  is  the  axis  of  revolution. 


290  INTEGRAL  CALCULUS 

A  derivation  as  the  limit  of  a  sum  has  been  given  in  Art. 
153.  In  the  figure  of  this  Art.  160,  if  V  is  the  volume  made 
by  the  revolution  of  the  area  M0P0PM  about  OX,  then  dV 
is  the  volume  of  the  cylinder  MPN,  whose  base  is  iry2  and 
whose  length  is  PN  =  dx.  This  is  so,  for  this  volume  is 
what  the  change  of  the  volume  V  would  be,  if  at  P  the  change 
became  uniform,  as  x  increased  uniformly  by  Ax  =  dx  from 
that  point.  As  in  the  case  of  the  surface  S  the  volume  V 
may  be  considered  as  being  generated  by  a  circle  of  varying 
radius  y,  the  center  of  the  moving  circle  always  on  the  x-axis 
and  the  point  P  moving  on  the  curve  according  to  its  equation. 

By  the  method  of  limits,  it  is  evident  that,  if  P'R  =  As, 

volume  MM'P'  >  A  7  >  M'MP; 
that  is,  7T  (y  +  Ay)2  Ax  >  A  V  >  iry2  Ax ; 

AV 

dividing  by  Ax,     t  (y  +  Ay)2  >  -r—  >  wy2; 

hence,     lim    -r—    =  -=-  =  iry2,  since  Ay  =  0,  as  Ax  =  0; 

.*.     dV  =  iry2  dx    or     V  =  w  I    y2  dx.  (6) 

If  the  revolution  is  made  about  a  line  y  =  b,  then 

V  =  rr  fX(y-b)2dx,  (8) 

and  when  the  revolution  is  about  a  line  x  =  a,  then 

(x-a)2dy.  (9) 


V  =7T    fy 


Note.  —  It  may  be  noted  that  the  cone  and  the  sphere  of 
Art.  155  and  the  paraboloids  of  Art.  156  and  Art.  159  are  all 
solids  of  revolution,  and  hence  the  formulas  of  this  Art.  160 
are  applicable  to  the  determination  of  their  surfaces  and 
volumes. 


SURFACES  AND  SOLIDS  OF  REVOLUTION 


291 


Example  1.  —  Find  the  volume  generated  by  the  revolu- 
tion of  the  area  of  the  equilateral  hyperbola  xy  =  1  about 
OX. 

V  =  it  I  y2dx  =  ir  I    —9dx=—7r   -      =tt ; 


ko=o 


=  oo; 


hence,  the  entire  volume  has  no  limit. 

r  1       llXo=1 

y  =  7r =7r  cubic  units; 

L^o      xjx=a0 


1M 


X 


hence  the  limit  of  the  volume,  from  the  section  at  x0  =  OM0 
=  1,  extending  indefinitely  to  the  right,  is  the  same  as  the 
volume  of  the  cylinder  generated  by  the  revolution  of  NP0, 
the  abscissa  of  P0,  about  OX.  Thus,  while  the  area  under 
the  curve  y  =  1/x,  from  the  ordinate  MqP0  at  x  =  1,  in- 
definitely to  the  right,  is  unlimited  (as  shown  in  Ex.  11,  Art. 
135),  the  volume  made  by  its  revolution  about  OX  has  a 
definite  limit.     According  to  Art.  156,  if  the  curve  y"=ir/x 


292  INTEGRAL  CALCULUS 

is  drawn,  any  one  of  its  ordinates  in  linear  units  will  represent 
the  volume  of  the  solid  extending  indefinitely  to  the  right  of 
that  ordinate;  thus,  in  the  figure  the  ordinate  M0Po"  =  tt 
represents  the  volume  to  the  right  of  P0MoP0',  and  the 
ordinate  MiPi"  =  J  w,  the  volume  to  the  right  of  PiMiPi. 
In  general,  the  ordinate  MP"  at  x  =  OM  represents  the 
volume  of  the  solid  to  the  right  of  the  section  at  any  distance 
x  from  the  origin,  and  it  represents  also  the  area  under  the 
curve  y  =  tt/x2  to  the  right  of  the  ordinate  to  that  curve. 

Example  2.  —  Find  the  volume  to  the  left  of  the  ?/-axis  of 
the  solid  generated  by  the  revolution  of  the  exponential  curve 
y  =  ex  about  the  z-axis. 

Jr*y=i  fx=0  7T        1°  7T 

y2  dx  =  tv  j         e2x  dx=  -e2x\      =  »  cubic  units. 

(See  Ex.  2,  Art.  130,  for  figure.) 

EXERCISE   XXXIH. 

In  these  examples,  a  segment  of  a  solid  of  revolution  means  the 
portion  included  between  two  planes  perpendicular  to  its  axis,  the  solid 
or  its  segment  being,  in  general,  &  frustum;  and  a  zone  means  the  convex 
surface  of  a  segment. 

1.  Find  the  area  of  a  zone  of  the  paraboloid  of  revolution  about  the 

x-axis.     y2  =  2  px,  the  plane  curve.       Ans.   5—  [(p2+yrf  —  (p2+2A>2)*]. 

6  p 

See   Ex.  2,  Art.  157,  where  p  =  f ,  y0  =  0. 

2.  Find  the  area  of  a  zone  of  the  ellipsoid  of  revolution  about  the 
x-axis;  that  is,  a  zone  of  the  prolate  spheroid.     Get  entire  surface. 

62 
y2  =  —  (a2  —  x2)  =  (1  —  e2)  (a2  —  x2),  where  e  is  the  eccentricity. 

y  ds  =  -  v  a2  —  e2x2  dx. 
a 

:.     S  =  2tt-   C  Va2  -  e2x2 dx 
a  Jx 

h  r     /—■> n  ■    «2  •     ,^1x 

=  ir  -  \  x  v  a2  —  e2x2  -\ sin-1  — 

a\_  e  a  J  T0 

The  entire  surface    =  2irb  [b  +  (a/e)  sin-1e]. 


SURFACES  AND  SOLIDS  OF  REVOLUTION         293 

The  surface  of  a  sphere  =  limit  2  irb  [6+  (a/e)  sin-1  e]  =  2ra[a +a]  =  4  ira2, 

since  for  circle  e  =  0,  limit    — — -    =  lim    - —     =  1 ;  a  =  b. 

e=o  L     e     J      e=o  LsinflJ 

3.   Find  the  area  of  the  surface  generated  by  the  revolution  of  the 
cycloid  about  its  base. 

Taking  the  parametric  equations  of  the  cycloid, 

x  =  a  (0  —  sin  0),  y  =  a  (1  —  cos0); 
dx  =  a  (1  —  cos 0)  dd,  dy  =  asmddd; 
ds  =  Vdx2  +  dy2  =  a  V2  (1  -  cos  0)  dd. 

64     „ 


3'° 


S  =  2w  (yds  =  2wa2  C*V2(l-cosd)3M  =  lQwa2Csm3(^d(^\  = 

4.  Find  the  surface  generated  by  revolving  the  catenary  about  the 
y-axis,  from  x  =  0  to  x  =  a.     Also  about  the  rr-axis. 

Here     y  =  %\<P+e~«),     ds  =  ^\ea -\- e~~a)  dx. 
S  =  2  7T  Cx  ds  =  7T  §  x  \ea  +  e    a)dx 

r     (i     _f\      r/f     _£\    ] 

=  ir  \  x  •  a  \ea  —  e    a )  —  a  J     \ea  —  e    a)  dx    ,  by  parts, 
=  *\ax  (<P  -  e~  a)  -  a2  \f  +  e~ «/      =2  7ra2  (1  -  e"1). 

About  x-axis:       *S  =  27r  J  yds  =  2k  J    -  \e°  +  e    °/    dx 

=  tt  1 1  (e^  -  e~  V  +  a  J    =  ^  (e2  -  e"2  +  4). 

5.  Find  tne  entire  surface  generated  by  revolving  the  hypocycloid 
about  the  x-axis.     x*  +  y*  =  a*  is  the  equation  of  the  curve. 

*-»*/»* -9-a»jr»G)l*-4"*j«,]."-T'* 

6.  Find  the  area  of  a  zone  of  the  surface  generated  by  the  tractrix 
revolving  about  the  x-axis.     (See  Art.  150.) 

S  =  2tt  f  y  ds  =  2-k  f  y  (-  ^)  =  2ira  [-?/]*  =  2tcl  {y«  -  y). 


294  INTEGRAL  CALCULUS 

7.   A  quadrant  of  a  circle  is  revolved  about  a  tangent  at  one  extrem- 
ity.    Find  the  area  of  the  curved  surface  generated. 

S  =  2tt  j  (a  -  x)  ds  =  2tt  J°  (a  -  x)  (l  +  -2Y  dx, 

when  tangent  to  x2  +  y2  =  a2  is  perpendicular  to  z-axis, 
a2  dx  Ca      x  dx 


r  r     a2dx      _      ra      xdx     ~| 
=  27r  l_Jo    Va2  _  X2      aJ0    Va2  -x2\ 

=  2 T  [a2  sin"1  -  +  a  Va2  -  x2T  =  7m2  (tt  -  2). 
L  a>  Jo 


8.  Find  the  volume  of  a  segment  of  the  prolate  spheroid,  and  the 
entire  volume.  Find  the  latter  to  be  two-thirds  the  volume  of  the  cir- 
cumscribed cylinder  of  revolution. 

Am.   ~  \a2  (x  -  xo)  -■£  (xz  -  xo3)l- 

9.  Find  the  volume  of  the  oblate  spheroid,  that  is,  the  ellipsoid  of 
revolution  about  the  minor  axis  which  is  on  the  ?/-axis. 

Find  the  volume  to  be  two-thirds  of  that  of  the  circumscribed  cyl- 
inder of  revolution. 

10.  Find  the  volume  of  the  paraboloid  made  by  x2  =  2  py  about  the 
?/-axis.     (Compare  Ex.  3,  Art.  159.) 

Find  the  volume  to  be  one-half  that  of  the  circumscribed  cylinder  of 
revolution. 

11.  Find  that  the  volume  of  the  solid  generated  by  revolving  an 
arch  of  the  cycloid  about  its  base  is  five-eighths  of  the  circumscribed 
cylinder. 

Here         V  =  2*   f  *'     ,  V* dy  or     V  =  to*  C*  (1  -  cosd)3dd. 

Jo      V2ay  —  y2  J* 

12.  Find  the  volume  generated  by  the  catenary  revolving  about 
the  x-axis,  from  x  =  a  to  x  =  —a.  Also  find  the  volume  by  the  area 
with  the  same  arc  revolving  about  the  ?/-axis. 

Here         V  =  *  j_Jj  \e"  +  e    ~a)  dx  =  51  \°-e«  +  2x  -  |e     aJ 

=  x (c2  +  4  ~ e_2)  =  8-83  a'"'- 

And  V  =  tt  f  °  x2dy  =  £  C  x2  \f  -  e~  a)  dx. 

J 0  &  J  0 


SURFACES  AND  SOLIDS  OF  REVOLUTION         295 

Integrating  by  parts  gives 

V  =\  \ax2\ea+e    a)  -2a2x\ea  -  e    a)  +  2  a2  \e*  +  e~  A 

=  ^  (e  +  5  e"1  -  4)  =  0.878  a3. 

13.  Find  the  volume  of  the  solid  generated  by  the  revolution  of  the 
tract rix  about  the  x-axis. 

y2 dx  =  -ir  I    Va2  -  ifydy  =  —■ 

Ja  6 

14.  Find  the  volume  generated  by  the  revolution  of  the  hypocycloid 
about  the  x-axis. 

V  =tt  ( y-dx  =t    C   (a1  -  xl)*dx  =  ^tto3. 
J  J -a  lOo 

15.  Find  the  volume  generated  by  the  revolution  about  the  ?/-axis 
of  the  equilateral  hyperbola  xy  =  1,  from  x  =  0  to  x  =  1. 

V  =  ir  |        x*dy  —  x  I  ~  =  —  -       =7r  cubic  units. 

(Compare  Ex.  1,  Art.  160.) 

16.  Find  the  volume  of  the  segment  of  the  solid  generated  by  the 
revolution  of  the  equilateral  hyperbola  x°-  —  y2  =  a2  about  the  z-axis, 
the  altitude  of  the  segment  being  a,  measured  from  the  vertex. 

Ans.  jiraz. 

17.  Find  the  volume  generated  by  revolving  about  either  axis  the 
part  of  the  parabola  x5  +  \p  =  a?  intercepted  by  the  axes. 

Ans.  T;  jra3. 

18.  Find  the  volume  of  the  solid  generated  by  the  quadrant  of  a 
circle  revolved  about  a  tangent  at  one  extremity. 

V  =  7rf\a  -  x)2dy  =Trf\a-  V^=7)'  dy  =  ^  (|  -  |) 

19.  Find  the  volume  generated  by  the  revolution  of  the  cissoid 

x3 
y2  =  - — — —  about  the  x-axis,  from  the  origin  to  x  =  a. 

Ans.  |7ra3(31og2  -2). 

20.  Find  the  volume  generated  by  the  revolution  of  the  cissoid  about 
its  asymptote  x  =  2  a.  Ans.  2jr2a3. 


CHAPTER  V. 

SUCCESSIVE   INTEGRATION.     MULTIPLE 
INTEGRALS.     SURFACES  AND  VOLUMES. 

161.  Successive  Integration.  —  As  the  inverse  of  succes- 
sive differentiation  there  is  successive  integration.  If  a 
start  is  made  with  a  function  y  =  f  (x),  considered  as  an  nth 
derived  function,  a  single  integration  gives  another  function, 
the  integral;  the  integration  of  this  function  gives  a  second 
integral,  and  so  on.  The  result  of  n  integrations  is  the  nth 
integral  of  the  given  function. 

In  Art.  140  on  Integral  Curves  successive  integration  was 
indicated,  and  in  Art.  141  the  process  was  employed  in 
application  to  beams.  For  successive  integration  with 
respect  to  a  single  independent  variable,  in  general;  let 

fi(x)=ff(x)dx,  (1) 

f*(x)=ffi(x)dx,  (2) 

h  (x)  =  jh  Or)  dx.  (3) 

Since  f2(x)  =   /  [fi(x)]dx, 

it  follows  from  (1)  that 

ft(x)=f[ff(x)dx]dx;  (4) 

and  since  /3  (x)  =   I  [f2  (x)  ]  dx, 

296 


SUCCESSIVE  INTEGRATION  297 

it  follows  from  (4)  that 

fs  W  =  / 1  /  [ff  {X)  dX\  dX  \  dX'  (5) 

The  integral  in  (4)  is  called  a  double  integral  and  is  written 


ff> 


Similarly,  the  integral  in  (5)  is  called  a  triple  integral  and  is 
written 


/// 


f{x)  dx\ 

If  an  integral  is  evaluated  by  two  or  more  successive  in- 
tegrations, it  is  called  a  multiple  integral. 

For  example,  to  evaluate  the  multiple  integral  /    /    /  exdx?; 

ex  +  Ci  is  the  first  integral, 

ex  +  Cix  +  C2  is  the  second  integral, 

C  x2 
ex  +  —jr-  +  Cix  +  C3  is  the  third  integral. 


iff 


ix*  =  ex  +  ^f  +  C2x  +  C3. 


If  limits  are  given  for  each  successive  integration,   the 

integral  is  definite;  if  limits  are  not  given,  it  is  indefinite. 

d2s 
Example  1.  —  Given  the  acceleration  -r^  =  —  g  to  find  s. 

This  is  Ex.  5  of  Art.  115,  and  may  be  written  thus: 

s  =  I  (  —  gt+vo)  dt,  where  v0  is  the  constant  of  integration, 

s  =  —  \  gt2  +  vat  -\-  So,  where  sQ  is  the  constant  of  integration. 

Example  2.  —  Determine  the  curve  for  every  point  of 
which  the  rate  of  change  of  the  slope  is  2. 


298  INTEGRAL  CALCULUS 

Here  ^y  =  A.(^y)  =  ^  =  2- 

dx2      dx  \dx)       dx         ' 
.'.     y=Jj2dx\ 

y  =  I  (2x  +  Ci)  dx,  where  2  x  +  d  is  the  first  integral, 
y  =  x2  +  CiX  +  d,  the  second  integral. 

This  is  the  equation  of  any  parabola  that  has  its  axis  parallel 

to  the  y-axis  and  drawn  upwards,  and  its  latus  rectum  equal 

to  1 .     All  such  parabolas  may  be  gotten  by  giving  all  possible 

values  to  C\  and  Ci,  the  arbitrary  constants  of  integration. 

d2v 
Example  3.  —  Determine  the  locus  of  the  equation  -~  =  0. 

y=ffodx2, 

y  =  I  mdx,  where  m  is  the  constant  of  integration, 

y  =  mx  +  b,  where  b  is  the  constant  of  integration. 

The  locus  is  the  system  of  straight  lines,  the  arbitrary  con- 
stants m  and  b  representing  the  slope  and  ?/-intercept, 
respectively. 

Example  4.  —  In  the  theory  of  flexure  of  beams 

where  E,  I,  M,  R,  and  w  are  constants.  Get  an  expression  for 
y  and  determine  the  constants  of  integration  from  the  con- 
ditions, y  =  0  when  x  =  0,  and  y  =  0  when  x  =  I. 

»~TlJ[Mx+-2---6-  +  Cl)dx' 

1    [Mx2    .   Rx*       WX4    ,  „      t  „~\  n       .     n      n 


SUCCESSIVE   INTEGRATION  299 

n       1  [Ml2  ,  RP      id*  ,  r  71      .  . 

0  =  El [~Y  +  "6"  "  24  +  U\ '  when  X  =  Z; 

■"'        '      £/|_       2         6   +24_T 

J_  \Mtf      TW  _wtf  _m  _IW      wF\ 
y      £/[  2     +    6         24         2         6   +  24  J 

r2  r3  n 

Example  o. —  Evaluate  III    x3dx3. 

Jo    J  I    J 2 

Letting  I  denote  the  integral  and  making  the  integrations 
in  order  from  right  to  left; 

t        P  PM4^  2       «n   P  Pa  2   ' 

=  60  r  [YP  dx  =  120  pdx  =  240. 

Example  6.  —  A  point  has  an  acceleration  expressed  by 
the  equation  at  =  —  roo2  sin  cot,  where  r  and  a;  are  constants. 
Get  expressions  for  the  velocity  and  the  distance  or  space 
passed  over. 

Here    at  =  -7-7  =  —  roo2  sin  cot  and  s  =  I    I  —  rco2  sin  cot  dt2, 

ds       Cd2s  u  .f.      .j.        +-       ..„ 

. .     v  =  -j  =  I  -=■=■  dt  =  —  roo2  J  sin  oot  dt  =  roo  cos  cot  +  Ci, 

s  =   /  -r  dt  =  roo  I  cos  co£  eft  -f  /  Ci  dt, 

s  =  r  sin  oot  +  &£  +  C2, 
which  is  the  law  of  simple  harmonic  motion.     (See  Art.  73.) 

EXERCISE   XXXIV. 

1.  Evaluate  f  jf  (x2  -  1)  dx3.      Ans.  £  -  |  +  ^  +  C2x  +  C* 

2.  Evaluate  J  J  f  j  ^dx*- 

Ans.   - 1  log  x  +  J  x3  +  *  C>x2  +  C3x>  C4. 


300  INTEGRAL  CALCULUS 

3;  Evaluate  \   j   (   (  sinaxdx4. 

Ans.  1/a4  sin  ax  +  £  Gx3  +  \  C2x2  +  Csx  +  Ct. 

4.   Evaluate  f    f    f  x*dx*.  Ans.  16. 

Ji  J\  Jo 

6.   Evaluate  f    C  (  x3  dxK  Ans.  80. 

6.  Find  the  curve  at  each  of  whose  points  the  rate  of  change  of  the 
slope  is  four  times  the  abscissa,  and  which  passes  through  the  origin 
and  the  point  (2,  4).  Ans.  3  y  =  2  x  (x2  -  1). 


7.   Evaluate 


f2  f     Cr sind d&.  Ans.  t(0-  a). 

Jo     J  a    Jo 

d?s 

8.  The  differential  equation  of  falling  bodies  is  -r=r  =  —  g]    show 

that  s  =  -  ~  +  Ci<  +  C2;  and  find  G  and  C2,  if  s  =  0  and  v  =  100, 

when  t  =  0. 

9.  A  point  has  an  acceleration  expressed  by  the  equation  at  = 
— rco2  cos  cot,  where  r  and  co  are  constants.  Find  expressions  for  the 
velocity  and  the  distance  passed  over.  Find  G  and  C2,  if  s  =  r  and 
v  =  0,  when  t  =  0. 

Ans.  v  =  —rco  sin  ut  +  G;  s  =  r  cos  coi  +  G£  +  Cs 

162.  Successive  Integration  with  Respect  to  Two  or 
More  Independent  Variables.  —  In  the  preceding  Article 
successive  integration  was  of  functions  with  respect  to  a 
single  independent  variable.  Successive  integration  of 
functions  of  several  independent  variables  are  now  to  be 
considered.  Suppose  there  is  given  a  function  /  (x,  y,  z)  of 
three  independent  variables. 

Let  /,  (x,  y,  z)  =  J  f  (x,  y,  z)  dz,  (1) 

h  fo  V,  z)  =  J  /i  0,  V,  z)  dy,  (2) 

/s  fa  V,  z)  =  J  h  fa  V,  z)  dx,  (3) 

where  in  (1)  the  integration  is  with  respect  to  z,  that  is,  as  if 
x  and  y  were  constants.     Likewise  in  (2)  it  is  with  respect 


THE  CONSTANT  OF  INTEGRATION  301 

to  y,  as  if  x  and  z  were  constants,  and  in  (3)  with  respect  to 
x,  as  if  y  and  z  were  constants. 

Equation  (2),  by  substitution  from  (1),  becomes 

h fo  y,z)=j  \Jf  fe y, «) <fc| fty;  (4) 

and  equation  (3),  by  substitution  from  (4),  becomes 

f3(x,  y,  z)  =  J  \  J  \J  (x,  y,  z)  dz\  dy  I  dx.  (5) 

The  integral  in  (5)  is  called  a  triple  integral  and  is  written 


/// 


f(x,  y,  z)dxdydz,  (6) 

where  the  order  of  the  integrations  is  from  right  to  left;  that 
is,  the  differential  coefficient  /  (x,  y,  z)  is  to  be  integrated 
with  respect  to  g,  that  result  to  be  integrated  with  respect  to 
y,  and  finally  the  last  result  is  to  be  integrated  with  respect 
to  x. 

Similarly,  the  double  integral  in  (4)  is  written : 


// 


f{x,  y,  z)dydz. 

As  to  the  integration  signs,  the  first  on  the  right  is  to  be  taken 
with  the  first  differential  on  the  right,  which  is  dz  in  (6),  the 
second  sign  from  the  right  with  the  second  differential  from 
the  right,  and  so  on. 

If  when  limits  of  integration  are  given,  .they  are  constant 
limits,  the  order  of  the  integrations  may  be  reversed  without 
affecting  the  result,  but  when  the  definite  integral  has  variable 
limits  the  order  of  the  integrations  can  be  changed  only  by 
new  limits  adapted  to  the  new  order.  In  practical  problems 
the  limits  for  one  variable  are  often  functions  of  one  or  more 
of  the  other  variables. 

163.  The  Constant  of  Integration.  —  The  evaluation  of 
an  indefinite  multiple  integral  differs  from  that  of  an  indefi- 
nite single  integral  in  the  form  of  the  constant  of  integration. 


302  INTEGRAL  CALCULUS 

Thus  I  /  ixy  dx  dy  being  given,  to  find  a  function  u  of  x 
and  y  such  that 

cSydxdy  =  4xy 
is  the  problem.     It  is  evident  the  operations  represented  by 
,    ,   dx  dy  must  be  reversed  in  order  to  get  u. 

That  is,    u  =    /    /    ,    ,   dxdy  =   f    j  4xydx  dy,  (1) 

which  indicates  two  successive  integrations,  the  first  with 
respect  to  y,  x  and  dx  regarded  as  constants,  and  the  second 
with  respect  to  x,  y  being  regarded  as  constant.  Hence  the 
first  integration  gives 

-7-  =  2  xy2  +  constant  of  integration. 

Since  x  was  regarded  as  constant  during  the  integration, 
the  constant  of  integration  may  depend  upon  x,  that  is,  it 
may  be  some  function  <f>(x),  or  it  may  be  simply  C.  This  is 
so,  since  differentiating  either  2  xy2  +  C,  or  2  xy2  +  <f>  (x), 
with  respect  to  y  gives  the  same  result,  4  xy.     Hence, 

g  =  2a^  +  *(s), 

where  <f>  (x)  is  an  arbitrary  function  of  x  and  may  be  a  con- 
stant C. 

Integrating  this  result,  with  y  constant,  gives 

u  =  xhf  +  f<f>  (x)  dx  +  F  (y),  (2) 

where,  since  y  was  regarded  as  constant  during  the  integra- 
tion, the  integration  constant  is  an  arbitrary  function  of  y 
and  may  be  C  with  a  constant  value,  possibly  zero. 


THE  CONSTANT  OF  INTEGRATION  303 

By  referring  to  Art.  109,  (2),  it  will  be  seen  that  if 
u  =  x3  +  x2y2;     -r—j-dxdy  =  4:xydxdy; 

that  is,  4>(x)  =3  x2  and  F  (y)  =  0,  for  that  function  u  of 

0,  y). 
The  indefiniteness  of  the  result  in  (2)  is  manifest,  for 


u  =   I    I  \xydxdy  =  x2y2, 


if  both  constants  of  integration  are  zero,  that  is,  <£  (x)  =0 
and  F  {y)  =  0.     The  indefiniteness  is  removed  when  limits 
for  the  variables  are  given,  the  integral  being  then  a  definite 
integral. 
Example.  — 

I      I    xyz dxdydz=   I       I    xydxdy\-\ 

a       Jo     Jy  Ja       t/0  |_~  Jv 

EXERCISE   XXXV. 

Evaluate  the  following  integrals : 

1.    ff#V  <b  dy.  Arts.  \  x3y*  +  F  (x)  +  h  (y). 


-  Va*-x* 


2.  (      I  x2ydxdy. 

Jo    Jo 

3.  C  f    p2  sin  6  dp  dd  =  5- 

p 

4.  C    f  pdpdd  =  &  b2. 

I       xydydx  =  \\b\ 
0   Jy—b 


rz362 
An..    35. 


304  INTEGRAL  CALCULUS 


J"*    /»2aco90  tt 

f  Pd9dP=l(a>-V). 

0     ^2bcosd  6 

Xb   fy  53  —  O? 

I    p2  sin  Bdpdd  =  — ^—  (cos  /3  —  cos  7). 

^*   J    J      ^~*2^ds  =  6&3. 

J«3     /»2     /»5 
I     J    xy2dxdydz  =  17£. 
2    •'1     •'2 
/»3     /»2     fh 

10.  J     I     I   x?/2  dz  dy  dx  =  2\\. 

11.  J*  J*  f*xy*dzdy  dx  =  17%. 

12.  f     f  f  *  x>y*z  dx  dydz  =  \  a?V{b*  -  a3). 

J  a    Jo    Jb 

J~2   rx  rx+y  e8  —  3      3  e* 

0    ./o   »/o  o  4 

14.     fb  fh  V2jy  dxdy  =  l  ^¥g  (h2*  -  hj)  b. 

Jo   J  hi 


4*    r1*  ra  JrtJ     a2(8-7r) 

15.     I      I  pdddp  =  — ^ -■ 

Jo    J  a(l-cos0)  o 


16.     f      f  °  (w  +  2v)dvdw  =  W  ^. 
I     I      x2  2/2  dx  dy  dz  =  32  a7. 

0       Jo    Jly 

164.  Plane  Areas  by  Double  Integration  —  Rectangular 
coordinates.  —  It  has  been  shown  in  Art.  135,  that  the  area 
between  two  curves  y  =  f  (x)  and  y  =  F  (x)  is  given  by 

A=    r\f(x)-F(x))dx,  (1) 

t/x0 

where  the  points  of  intersection  are  (x0,  y0)  and  (xh  y\).  The 
area  is  thus  given  not  by  a  single  integral  but  by  the  differ- 
ence between  two  integrals,    /    f(x)dx  —   I    F  (x)  dx.     The 

«A0  «/x0 

result  is  gotten  also  by  double  integration,  finding  the  limit 
of  two  sums.  Let  the  element  of  area  be  ly  lx,  (x,  y)  being 
any  point  P  of  the   area.     If  the  elements  are  summed 


PLANE  AREAS  BY   DOUBLE  INTEGRATION 


305 


up  with  respect  to  y,  with  the  limits  MD  and  MN,  or  F(x) 
and  /  (x),  x  being  constant,  the  area  of  the  strip  DN'  is 
gotten.  If  the  strips  are  summed  up  with  the  limits  a  and 
b  for  x,  then 

V    %±x\Ay=2    %AxAy 

x=a\_F(x)  J  aF(x) 

is  the  expression  for  the  sums.     Taking  the  limits  of  the 


sums,  first  as  Ay  =  0  and  then  as  Ax  =  0,  the  area  ADBN 
is  given  by  the  double  integral 

nf{x) 
dxdy,  (2) 

(x) 

which  integrated  first  with  respect  to  y  gives 

A=  £u(x)-F(x))dx.  (1) 

If  the  elements  are  summed  up  in  reverse  order,  first  with 
respect  to  x  with  the  limits  H'H  and  H'S,  or  /-1  (y)  and 
F^iy),  y  being  constant,  and  then  with  respect  to  y  with 
limits  c  and  d,  there  results 

dydx,  (3) 

.  Hv) 


306  INTEGRAL  CALCULUS 

where/-1  (y)  and  F_1  (y)  are  the  inverse  functions  of  f(x)  and 
F  (x),  respectively. 

Integrating  (3)  the  area  ADBN  is  gotten,  as  given  by  (2). 
Hence,  in  general, 

A=   f  fdx  dy  (4) 

is  the  formula  for  area  by  double  integration,  the  limits  being 
taken  so  as  to  include  the  required  area.  The  order  of 
integration  is  indifferent  provided  the  limits  be  adapted  to 
the  order  taken. 

Corollary.  —  dxy2A  =  dx  dy  and  dyx2A  =  dy  dx. 

Example  1.  —  Find  the  area  bounded  by  the  parabolas 
y2  =  2  px  and  x2  =  2  py. 

The  parabolas  intersect  at  the  points  (0,  0)  and  (2  p,  2  p). 

J^2P   rVjpl 
I         dxdy  =  f  p2,  by  formula  (2) . 
0  J  12 

f 

dy  dx  =  f  p2,  by  formula  (3). 


■2p 
2py 


Example  2.  —  Find  the  area  bounded  by  the  circle  x2  +  y2 
=  12,  the  parabola  y2  =  4  x,  and  the 
parabola  x2  =  4  y. 

For  the  part  OPP2  the  limits  for  x  are 
0  and  2,  while  for  the  part  PP1P2,  they 
are  2  and  Vs,  the  point  P  being  (2,  Vs) 
and  the  point  Px  (Vs,  2).  For  both  parts 
the  lower  limit  for  y  is  the  ordinate  of 
x2  =  4  y;  for  OPP2  the  upper  limit  for  y  is  the  ordinate  of 
y2  =  4  x,  and  for  PPiP2)  that  of  x2  +  y2  =  12. 

A  =  OPPiP*  =  OPP2  +  PPiP-2  =    (2  [''dxdy 

Jo    «7/2 
4 
/»\8     /»v'i2-j2 

+   /        /  dxdy  =  3.92. 

«/2         t/j* 


PLANE  AREAS  BY  DOUBLE  INTEGRATION        307 

Example  3.  —  Find  the  area  between  the  parabola  y2  =  ax 
and  the  circle  y2  =  2  ax  —  x2. 


nv2az-z2 
dxdy 
'ax 

Jo  Z  6 

Note.  —  It  may  be  seen  that  in  finding  some  areas  there  is 
no  advantage  in  using  double  integration,  as  after  the  first 
integration  with  the  limits  substituted,  the  remaining  in- 
tegral is  what  might  have  been  formed  at  first.  There  are, 
however,  cases  where  double  integration  furnishes  the  only 
method  of  solution;  hence  the  need  for  some  practice  in  its 
application. 


EXERCISE   XXXVI. 

1.   Find  the  area  between  the  circle  x2  +  y2  =  a2  and  the  line  y 


A-   f   f^dzdy-'-^*. 

•'o     Ja—x  4 

2.  Find  by   double  integration  the  area  between  the  parabolas 
y2  =  8  x  and  x2  =  8  y. 

Ans.  21§. 

3.  Find  the  area  bounded  by  the  circle  x2  +  y2  =  25,  the  parabola 
y*  =  -L6  x,  and  the  parabola  y  =  t&  x2. 

Ans.  7.55. 

4.  Find  by  double  integration  the  area   of  the 

x2       v2 
ellipse  -2  +  £-2  =  1. 

5.  Find  the  area  of    any  right    triangle,   using 
double  integration. 


a 

Ch  fk    j   j         Cha    j        a  x2lb      i 


ah. 


-x+a 


A  -  /*/  dxdy  =  £(-  %z  +  a)dx  =  -£  f  +  ax] *  =  \ ab. 


308 


INTEGRAL  CALCULUS 


165.   Plane  Areas  by  Double  Integration  —  Polar  coor- 
dinates.—  As  has  been  shown  in  Art.  135(b),  the  area  in 

polar  coordinates  of  P\OP2, 
generated  by  the  radius 
vector  p  as  0  increases  from 
0i  to  02  is  given  by 


-*£■ 


de. 


(i) 


To  find  the  area  between 
two  polar  curves  by  double 
integration,  let  the  element 
of  area  be  PDD'P',  bound- 
ed by  the  two  radii  OP', 
OD',  and  the  two  circular  arcs,  concentric  at  0. 

Let  the  coordinates  of  P  be  (p,  0) ;   then  from  geometry, 

sector  POD    =  \  p2  A0, 
sector  P'OD'  =  \  (p  +  Ap)2  A0. 

Hence,     A  A  =  PDD'P'  =  i  (p  +  Ap)2  AS]-  J  p2  A0 
=  (p  +  iAP)A0Ap. 

Keeping  A0  constant  and  summing  the  elements  of  area 
with  respect  to  p  gives  an  area  AA'B'B,  expressed  by 

OA  r*QA 

A0  •  lim  y.  (p  +  |  Ap)  Ap  =  A0   /       pdp. 

Ap=0  oa>  Jo  A' 

Making  the  summation  now  with  respect  to  0,  the  sum  of 
the  radial  slices  is  gotten,  and  the  limit  of  this  sum  is 

_.02  roA  re2   roA 

A  =  lim  V    A0-    /      Pdp=    I      I       pdpdd. 

A0=O  ^0y  J AO'  Jo,     JOA' 


Replacing  OA'  and  OA  by  F(d)  and  /  (0)  respectively,  the 

formula  is 

£"..    rue) 
I       pdddp.       -  (2) 

.    J  F  (0) 

When  F  (0)  =  0,  the  area  PiOP2  between  the  curve  p  =  /  (0) 


PLANE  AREAS  BY  DOUBLE  INTEGRATION        309 


and  the  radii  is  (1),  A  =  \ 


p2  dd,  where  (2)  has  been  in- 


tegrated as  to  p. 

If  the  summing  of  the  elements  of  area  be  made  first  with 
respect  to  0,  keeping  Ap  constant,  RSDP,  a  segment  of  a 
circular  ring,  is  gotten.  A  second  summation  with  respect 
to  p  gives  the  sum  of  such  ring  segments,  the  limit  of  which 
sum  is  the  area  A.     The  resulting  formula  is 


JPl     t/F-i(p) 


dpdd, 


(3) 


where  F_1(p)  and/_1(p)  are  the  inverse  functions  of  F(0)  and 
/  (0),  respectively. 

Corollary.  —  ddp2A  =  pdddp  and  dp02A  =  pdpdd  are  rec- 
tangles with  sides  p  dd  and  dp. 

Example  1.  —  A  simple  case  of  the  application  of  the 
formulas  is  in  finding  the  area  of  the  circle  p  =  a. 


(2) 


(3) 


Jr»2ir     Pa  P27T  ~~la 

/       pdOdp  =    \     /         p'ddl 
0      t/0  «/o  Jo 

=  ia2^T7r  =  7ra2 

n2ir  Pa  "]2tt 

pdpdd  =    /    pdpdl 
Jo  Jo 

=  27r^T  =  7ra* 


310 


INTEGRAL  CALCULUS 


In  (2)  the  sectors  are  summed,  while  in  (3)  the  rings  are 
summed.     In  this  case  of  the  circle  it  is  to  be  noted  that  no 

limit  need  be  invoked, 
since  the  integral  is  the 
sum  in  each  case,  the  in- 
crements being  the  differ- 
entials, the  variables  all 
*    increasing  uniformly. 

Example  2.  —  Find  the 
area  between  the  two  tan- 
gent circles  p  =  2  a  cos  0 
and  p  =  2  b  cos  0,  where 
a  >  b. 


TT  7T 

JP*     /*2acos0  /»* 

/  pdddp  =  4t(a2-b2)  I    cos2 6 

0    */2bcos0  *Jo 


dd 


=  ir{a2-b2). 

Example  3.  —  Find  the  areas  between  the  cardioid  p 
2  a  (1  —  cos  0)  and  the  circle  p  =  2  a. 


pdOdp 
_a(l-cos0) 

=  4  a2   /[I  -  (1  -  cos0)2]c?0 

7T 

=  4a2J     (2cos0-cos20)d0  =  8a2-7ra2. 


AREA  OF  ANY  SURFACE  BY  DOUBLE  INTEGRATION   311 

JrV     /»2«(l-cos0) 
/  pdddp 

w    nJ2a 
2 

=  4a2   P[(l  -cos0)2-  l]dd 
\ 

«=4a2  J(-2cos6  +  cos2e)dd  =  8a2+Tra2. 
I 

EXERCISE   XXXVH. 

1.  Find  by  double  integration  the  entire  area  of  the  cardioid  p  = 
2a(l— cos0).  Ans.  6?ra2. 

2.  Find  the  area  (1)  between  the  first  and  the  second  spire  of  the 
spiral  of  Archimedes  p  =  ad;  (2)  between  any  two  consecutive  spires; 
(3)  the  area  described  by  the  radius  vector  in  one  revolution  from  0  =  0, 
and  the  area  added  by  the  nth  revolution. 

Ans.  (1)  *#■ Tr*a2;  (2)  (n2  +  2n+  f) tH^2;  (3)  f  7r3a2,  f  (n3  -  1)  tt3^. 

3.  Find  by  double  integration  the  area  of  one  loop  of  the  lemniscate 
p2  =  a2  cos  2d.  Ans.  \  a2. 

4.  Find  by  double  integration  the  area  between  the  circle  p  =  cos  0 
and  one  loop  of  the  lemniscate  p2  =  cos  2  6.  Get  the  area  between  the 
circle  and  the  line  6  «=  tt/4  and  then  between  that  line,  the  lemniscate, 
and  the  circle.  .         ir  —  2      -k  —  2.x  —  2 

166.   Area  of  any  Surface  by  Double  Integration.  —  Let 

the  surface  be  given  by  an  equation  between  the  rectangular 
coordinates,  x,  y,  z.  Let  the  equation  of  the  given  surface 
be 

z  =f(*,y)- 

Passing  two  series  of  planes  parallel,  respectively,  to  XZ 
and  YZ,  will  divide  the  given  surface  into  elements.  These 
planes  will  at  the  same  time  divide  the  plane  XY  into  ele- 
mentary rectangles,  one  of  which  is  P'Pi',  the  projection 
upon  the  plane  XY  of  the  corresponding  element  of  the 
surface  PP2. 

Let  x,  y,  z  be  the  coordinates  of  P  and  x  +  Ax,  y  +  Ay, 
z  +  Az,  those  of  P2,  x  and  y  being  independent;  then  P'M'  = 
Ax  and  P'N'  =  Ay.    The  planes  which  cut  the  element  PP2 


312 


INTEGRAL  CALCULUS 


from  the  surface  will  cut  a  parallelogram  from  the  tangent 
plane  at  P,  the  projection  of  which  on  the  plane  XY  is  P'P2' 
=  Ax  Ay,  the  same  as  the  projection  of  the  element  PP2. 
The  projection  is  the  product  of  the  area  of  the  parallelogram 


and  the  cosine  of  the  angle  made  by  the  tangent  plane  with 
the  plane  XY;    hence,  denoting  the  angle  by  7  and  the 
parallelogram  cut  from  the  tangent  plane  by  PT, 
area  PT  =  area  P'P2'  •  sec  7 
=  Ax  Ay  sec  7. 
As  Ax  and  Ay  approach  zero,  the  point  P2  approaches  the 
point  P,  and  the  areas  PT  and  PP2  approach  equality;  that 
is,  the  element  of  surface  approaches  coincidence  with  the 
parallelogram,  a  portion  of  the  tangent  plane  at  P;     hence, 

area  PP2  =  Axy2S  =  Ax  Ay  sec  7,  approximately; 
that  is, 

area  PP2  =  Axy2S  =  Ax  Ay  sec  7 ;  Um    ^^  \  =  sec  7 ;  (  21  j 
.*.     dxu2S  =  sec  7  •  dx  dy.  (1) 


AREA  OF  ANY  SURFACE  BY  DOUBLE  INTEGRATION      313 

F^Art.103(8)>8ec,4l+(|)V(|)^(o^^) 

hence  from  (1), 


//[■+©'+(£)' 


dx  dy,  (2) 


the  limits  being  so  taken  as  to  include  the  desired  surface. 

Let  £  denote  that  part  of  the  surface  z  =  f(x,  y),  z  being 

a  one-valued  function,  which  is  included  by  the  cylindrical 

surfaces  y  =  0O  (x),  y  =  4>  (x),  and  the  planes  x  =  a,  y  =  b; 

-    s-f •CHINS)"]'**    <*> 

In  finding  the  area  of  the  given  surface  a  more  convenient 
form  of  the  equation  of  the  surface  may  be  either  x  = 
f  (y,  z),  or  y  =  /  (z,  x).  The  formula  for  the  area  will  be 
then  either 


//['+(!)*+< 


dx" 
\dzj 


dydz,  (3) 


or 


//['+@H2)7^ 


with  the  proper  limits  of  integration. 

In  applying  the  formulas,  the  values  of  the  partial  deriva- 
tives are  gotten  from  the  equation  of  the  surface  the  area  of 
which  is  sought;  hence,  when  there  are  two  surfaces  each  of 
which  intercepts  a  portion  of  the  other,  the  partial  derivatives 
in  each  case  are  taken  from  the  equation  of  that  surface 
whose  partial  area  is  being  sought.  This  will  be  illustrated 
in  the  following  examples. 

Example  1.  —  To  find  the  surface  of  the  sphere  whose 
equation  is 

x2  +  y2  +  z2  =  a2. 


314  INTEGRAL  CALCULUS 

Let  0  -  ABC  of  the  figure  (Art.  166)  be  one-eighth  of 
the  sphere. 

dx~       z1        dy~       z* 

i-iVfeY.i  f*Y-i  i  *2  i  y2  =  a2=      a* 

1  "*"  \dx)  ~*~\dy)       l'r#~r#      z2      a2  -  x2  -  y2 

Area  =  8  =  8 ABC  =  8a  P  f  *         ,     ^^        by  (2) 

Jo  Jo  Va2-z2-?/2 

=  8  a  /  "sin-1     ,    y  ds 

Jo  Va2  -  x2Jo 

a        Ca a        a      2       /    Compare  Ex.  2,    \ 
=  4:ira  I    dx  =  4ira2.         ^        .      v  ' 

Jo  V  Exercise  XXXIII.  / 

Here  the  integration  was  over  the  region  OAB,  the  projec- 
tion of  the  curved  surface  ABC  on  IF  plane.  The  first 
integration  with  respect  to  y  summed  all  the  elements  in  a 
strip  LL'K'K,  y  varying  from  zero  to  NL',  that  is,  between 
limits  0  and  Va2  —  x2 ,  the  equation  of  the  intersection  of  the 
surface  with  the  XY  plane  being  x2  +  y2  =  a2.  Integrating 
next  with  respect  to  x,  the  surface  ABC  is  gotten  by  sum- 
ming all  the  strips  from  x  =  0  to  x  =  a. 

Example  2.  —  Find  the  area  of  the  portion  of  the  surface 
of  a  sphere  which  is  intercepted  by  a  right  cylinder,  one  of 
whose  edges  passes  through  the  center  of  the  sphere,  and 
the  radius  of  whose  base  is  half  that  of  the  sphere. 

Note.  —  This  is  the  celebrated  Florentine  enigma,  pro- 
posed by  Vincent  Viviani  as  a  challenge  to  the  mathemati- 
cians of  his  time.     (Williamson's  Integral  Calculus.) 

Taking  the  origin  at  the  center  of  the  sphere,  an  element 
of  the  cylinder  for  the  2-axis  and  a  diameter  of  a  right  section 
of  the  cylinder  for  the  x-axis,  the  equation  of  the  sphere  will 
be  x2  +  y2  -f-  z2  =  a2,  and  the  equation  of  the  cylinder, 
x2  _|_  y2  _  ax 


AREA  OF  ANY  SURFACE  BY  DOUBLE  INTEGRATION'  315 

The  area  of  APCD  is  one-fourth  of  the  area  sought,  and  since 
this  surface  is  a  portion  of  the  surface  of  the  sphere,  the  par- 

tial  derivatives  -7- ,  -j-  must  be  taken  from  x2  +  y2  +  z2  =  a2, 
ax  dy  v  ' 

giving,  as  in  Ex.  1,  using  formula  (2), 

S=    f  f       adxdy 
J  J   Va2  —  x2  —  y2 
to  be  integrated  over  the  region  OP' A.     Hence, 


The  limits  for  y  are  from  x2  +  y2  =  ax,  the  equation  of  the 
curve  OP' A,  the  boundary  of  the  projection  of  the  surface 
APCD  on  the  XY  plane. 

Example  3.  —  Find  the  surface  of  the  cylinder  of  Ex.  2, 
intercepted  by  the  sphere. 

The  area  of  APCOP'  is  one-fourth  of  the  area  sought, 
and  since  it  is  a  part  of  the  lateral  surface  of  the  cylinder 
x2  +  y2  =  ax,  the  partial  derivatives  in  formula  (2)  must  be 


316  INTEGRAL  CALCULUS 

dz 
taken  from  this  equation.     But  from  this  equation  ^  =  oo , 

—  =  oo,  and  formula  (2)  does  not  apply,  which  is,  more- 
dy 

over,  evident  since  the  element  of  surface  is  dx  dz  in  the  strip 
P'P,  and  the  area  of  the  surface  A  PC  OP'  cannot  be  found 
from  its  projection  on  the  XY  plane,  for  this  projection  is  the 
arc  AP'O.  The  projection  is  made  on  the  XZ  plane  and 
formula  (4)  used. 
The  partial  derivatives  are  found  to  be 

dy  =  a-2x      dy  =  Q 

dx  y      '    dz 

Since  P  is  on  the  sphere, 

pp2  =  z2  =  a2  -  (x2  +  y2)  =  a2  -  ax, 
since  P  is  on  the  cylinder.     Hence, 

dx  dz  rt      Ca  ^a2  —  cix 


~2«  r  f    /"*  =2«r 

Jo  Jo  V ax  —  x2  Jo 

=  2a  fa\/^dx  =  4:a' 


Vc 


dx 


Here  the  integration  is  over  the  region  0AP"C,  AP"C  being 
the  projection  of  A  PC  on  XZ  plane.  The  first  integration 
sums  up  the  elements  of  surface  in  the  strip  P'P  and  the 
next  integration  sums  up  the  strips  from  x  =  0  to  x  =  a. 

By  eliminating  y  from  x2  +  y2  +  z2  =  a2  and  x2  +  y2  =  ax, 
z>  =  a2  —  ax  (as  found  above),  which  is  the  equation  of 
AP"C1  from  which  the  limits  of  z  are  taken. 

EXERCISE   XXXVHI. 

Find  by  double  integration  the  areas  of  the  surfaces  given  in  the 
following  examples: 

1.  The  zone  of  the  sphere,  x2  -f-  if  +  z-  =  r2,  included  between  the 
planes  x  =  a  and  x  =  b.  A  us.  2  wr  (6  —  o). 


VOLUMES  BY  TRIPLE  INTEGRATION  317 

2.  The  surface  of  the  right  cylinder  x2  -f  z2  =  a2  intercepted  by  the 
right  cylinder  x2  +  y2  =  a2.     Compare  Ex.  8,  Art.  157.       Ans.  8  a2. 

3.  The  part  of  the  plane  -  +  t  +  -  =  1,  in  the  first  octant,  inter- 

a      o       c 

cepted  by  the  coordinate  planes.  Ans.  §V '  a2\f-  -f  a2c2  -+-  b2c2. 

4.  The  surface  of  the  cylinder  x2  +  y2  =  a2,  included  between  the 
plane  z  =  mx  and  the  XY  plane.  Find  by  both  formula  (3)  and 
formula  (4),  and  show  why  formula  (2)  does  not  apply. 

Ans.  4  ma2. 

5.  The  surface  of  the  paraboloid  of  revolution  y2  +  z2  =  4  ax, 
intercepted  by  the  parabolic  cylinder  y2  =  ax  and  the  plane  i  =  3a. 

6.  The  surface  of  the  cylinder  of  Ex.  5,  intercepted  by  the  parabo- 
loid of  revolution  and  the  given  plane. 

I  [   %r        rfx  d«  =  2  V3  (      (4  ax  +  a2)^  dx 

=  (13VT3-1)^. 

167.  Volumes  by  Triple  Integration  —  Rectangular  Co- 
ordinates. —  Let  the  volume  be  that  of  a  solid  bounded  by 
the  coordinate  planes  and  any  surface  given  by  an  equation 
between  the  coordinates  x,  y,  and  z. 

Let  P  be  any  point  (x,  y,  z)  within  the  solid  0  —  ABC,  the 
surface  being  given  by  z  =  f  (x,  y) ,  where  /  (x,  y)  is  a  con- 
tinuous function.  Let  K'  be  the  point  (x  +  Ax,  y  +  \y, 
z  +  Az),  the  diagonally  opposite  corner  of  the  rectangular 
parallelopiped  formed  by  passing  planes  through  P  and  K', 
the  planes  being  parallel  to  the  coordinate  planes.  Let  more 
planes  be  passed.  Taking  first  the  sum  of  the  elementary 
parallelopipeds  whose  edges  lie  along  the  line  NT,  the  limit 
of  this  sum,  as  A z  is  made  to  approach  zero,  is  the  volume  of 
the  prism  whose  base  is  Ix  Ay  and  whose  altitude  is  NT, 
x,  y,  Ax,  and  Ay  remaining  constant  during  the  summation. 
Next  with  x  and  Ax  constant,  sum  the  prisms  between  the 
planes  MHL  and  SDR.  The  limit  of  this  sum  as  Ay  is  made 
to  approach   zero  is  the  volume  of  the   cylindrical   slice 


318 


INTEGRAL  CALCULUS 


LR'D'HMS.  Finally,  when  taking  the  sum  of  the  slices 
parallel  to  the  YZ  plane,  as  Ax  approaches  zero,  the  volume 
of  the  cylindrical  slice  approaches  that  of  the  actual  slice 
LRDHMS;  hence,  the  limit  of  the  sum  of  the  slices,  as  Ax 
approaches  zero,  is  the  volume  of  the  solid. 


Jr*OA     PMH    (*NT 
I         I       dx  dy  dz, 
o      Jo       Jo 


where  V  is  the  volume  of  0  —  ABC.  Let  V  denote  the 
volume  bounded  by  the  curved  surfaces  z  =  f0{x,  y),  z  = 
f(x,  y);  the  cylindrical  surfaces  y  =  <f>o(x),  y  =  4>(x);  and 
the  planes  x  =  a,  x  =  b;  then 

I         dxdydz.  (1) 

.  j(j)   Jfu(x,y) 

Corollary.  — dxyzW  =  dxdydz,     dyJV  =  dydzdx,  .  .  . 
If  z  is  expressed  in  terms  of  x  and  y,  and  f0  (x,  y)  =  0; 


a      J,j)0(x) 


z  dx  dy. 


VOLUMES  BY  TRIPLE  INTEGRATION  319 

Note.  —  The  formula,  V  —  I  j  I  dx  dy  dz,  may  be  de- 
rived from  the  figure  by  the  definition  of  differentials. 

Thus,  the  variables  x,  y,  z,  being  independent,  dx,  dy,  dz 
may  be  taken  as  finite  constants,  the  parallelopiped  PK'  being 
dx  dy  dz.  When  x  and  y  are  regarded  constant,  PK'  is  the 
differential  of  the  prism  NK.  Hence,  integrating  dx  dy  dz 
between  the  limits  z  =  0  and  z  =  NT  gives  the  prism 
NT  dx  dy,  which  is  the  differential  of  the  solid  MSR'L  -  T. 
Integrating  NT  dx  dy  between  the  limits  y  =  0  and  y  =  MH 
gives  the  cjdinder  MLH  —  D' ',  or  MLH  dx,  which  is  the  differ- 
ential of  the  solid  OBC  —  M.  Integrating  MLH  dx  between 
the  limits  x  =  0  and  x  =  OA  gives  the  volume  OBC  —  A,  or 

V.     Hence,  TT  =  /    l    f  dx  dy  dz,  the  limits  being  so  chosen 

as  to  include  the  volume  sought. 

Example.  —  Find  the  volume  of  the  ellipsoid 

t  +  yl  +  t=  ! 

az      oz      c 

The  entire  volume  is  eight  times  that  in  the  first  octant, 
where  the  limits  are: 


z  =  0,     z  =  cVl  -  x2/a2  +  y2/b2; 
y  =  0,    y  =  bV\  -  x2/a2; 
x  =  0,     x  =  a; 


/  dxdydz  = 

Jo 
Corollary.  —  For  sphere,  a  =  b  =  c;     .'.     1 


3 

4  7ra3 


/  dy  dx  dz 

*Jo 

naVl_02/c2       />bVl_X2/a2_22/c2 
/  dz  dx  dy. 

Jo 


320  INTEGRAL  CALCULUS 

EXERCISE   XXXIX. 

Find  by  triple  integration  the  volumes  required: 

1.  The  tetrahedron  bounded  by  the  coordinate  planes  and  by  the 
plane 

a      b       c  o 

See  Ex.  3,  Exercise  XXXVIII,  for  the  surface  of  the  plane. 

2.  The  volume  bounded  by  the  cylinder  x2  +  y2  =  a2  and  the  planes 

.        4  ma3 
z  =  0  and  z  =  mx.  Ans.  — — 

3.  A  cylindrical  vessel  with  a  height  of  12  inches  and  a  base  diam- 
eter of  8  inches  is  tipped  and  the  contained  liquid  is  poured  out  until 
the  surface  of  the  remaining  liquid  coincides  with  a  diameter  of  the 
base.     Find  the  volume  remaining  in  the  vessel.         Ans.  128  cu.  in. 

Note  that  the  volume  is  one-half  that  given  by  Ex.  2,  above. 

4.  The  volume  included  between  the  paraboloid  of  revolution 
yi  +  z2  =  4  ax,  the  parabolic  cylinder  y2  =  ax  and  the  plane  x  =  3  a. 
See  Exs.  5  and  6,  Exercise  XXXVIII,  for  the  surfaces. 

J-3a   /"Vax    /»(4ox-y2)^  ,  ,-N 

(  dxdydz  =  {Gir  +  9V3)a3. 

o     J  o        «A) 

5.  The  volume  included  between  the  paraboloid  of  revolution 
xt  -\-yi  =  az,  the  cylinder  x2  +  y2  =  2  ax,  and  the  XY  plane. 

.      x*+& 

Xia    pV^ax-x*    r      a 
I  I  dxdydz  =  f  ira3. 


6.   The  entire  volume  bounded  by  the  surface 


Ans-V=Ll  I  dxdydz -w 

7.  The  entire  volume  bounded  by  the  surface 

?    i     i    .     %        i  a         4  ira3 

x s  +  ?/ J+z s  =  a!.  Ans.   -qH-* 

8.  The  volume  of  the  part  of  the  cylinder  intercepted  by  the  sphere. 
The  radius  of  the  sphere  is  a  and  it  has  its  center  on  the  surface  of  a 
right  cylinder,  the  radius  of  whose  base  is  a/2.  See  Exs.  2  and  3, 
Art.  166.  

J -a    r\/ax-x*   /•Va»— »«— «l 
I  I  dxdydz  =  i(ir-  f)a». 


SOLIDS  OF  REVOLUTION  BY  DOUBLE  INTEGRATION     321 

168.  Solids  of  Revolution  by  Double  Integration.  —  In 
the  figure  of  Art.  164,  where  P(x,  y)  is  any  point  in  the  area 
ADBN,  x  and  y  being  independent,  Ax  Ay  is  the  element 
of  area.  Conceive  the  area  ADBN  to  revolve  through  0 
radians  about  OX  as  an  axis;   then 

By  ■  Ax  ly  <  AxyW  <  0  {y  +  Ly)  ■  Ax  Ay; 

■■  ey<tS-y<d(y+^'> 

hence,"  hm  \     \=  -r- j-  =  dy; 

AX,Ay=olAxAyj     dxdy 

.'.     d2V  =  By  dxdy, 

/       ydxdy.  (1) 

Similarly,  about  OY, 

Jrvi  rrny) 
I         x  dy  dx.  (2) 

2/0    JF-Hy) 

Putting  6  =  2  7r,  the  formulas  give  the  volumes  generated 
by  a  complete  revolution  of  the  area. 

Corollary.  —  If  the  axis  of  revolution  cuts  the  area,  (1) 
or  (2)  will  give  the  difference  between  the  volumes  generated 
by  the  two  parts.  Hence  7  =  0,  when  these  two  parts 
generate  equal  volumes.  Integrating  (1)  first  with  respect 
to  y,  and  (2)  first  with  respect  to  x,  the  upper  limits  being 
the  variables  y  or  x  and  the  lower  limits  zero,  gives 

V  =  ir  fXly2dx  (lr) 

and  V  =  tt  fmx2dy,  (2') 

the  formulas  for  solids  of  revolution,  single  integration. 

169.  Volumes  by  Triple  Integration  —  Polar  Coordinates. 
—  Let  the  point  P  (p,  6,  <f>)  be  any  point  within  a  por- 
tion of  a  solid  bounded  by  a  surface  and  the  rectangular 
planes.  As  usual,  p  is  the  distance  OP  from  the  pole  at  the 
origin,  6  is  the  angle  ZOP  which  OP  makes  with  the  z-axis, 


322 


INTEGRAL  CALCULUS 


and  <f>  is  the  angle  XOP'  which  the  projection  of  OP  on  the 
XY  plane  makes  with  the  x-axis.  Let  the  solid  be  divided 
into  elementary  volumes   like  PDD1Q1   by   the  following 


means. 


(1)  Through  the  2-axis  pass  a  series  of  consecutive  planes, 
dividing  the  solid  into  wedge-shaped  slices  such  as  COAB. 

(2)  Round  the  2-axis  describe  a  series  of  right  cones  with 
their  vertices  at  0,  thus  dividing  each  slice  into  elementary 
pyramids  like  0  -  RSTV. 

(3)  With  0  as  a  center  describe  a  series  of  consecutive 
spheres.  Thus  the  solid  is  divided  into  elementary  solids 
like  PDDiQi,  whose  volume  is  given  approximately  by  the 
product  of  three  of  its  edges,  PPlf  PPi,  and  PQ. 

Let  edge  PQ  =  Ap,  angle  POP2  =  A0,  angle  AOB  =  angle 
PO'Pi  =  A(j>;  then  edge  PPX  =  p  sin  0  A</>,  and  edge  PP2  =» 
PA0. 

Hence,  the  volume  of  the  elementary  solid  is  given  ap- 
proximately   by    p2  sin  0  A0  A<£  Ap.     It    can    be    expressed 


VOLUMES  BY  TRIPLE  INTEGRATION  323 

exactly  but  the  additional  terms  vanish  when  the  three  in- 
crements are  made  to  approach  zero.  Therefore,  the  volume 
of  the  solid  is  given  by  the  limit  of 

2X2  P2sin0A0A0AP; 

A0=O  A0=O  Ap=0 


■SIS 


p2  sin  0  dd  d<f>  dp,  (1) 

each  integral  to  be  taken  between  the  limits  required  to  find 
the  volume  sought.  The  summation  can  be  made  in  any 
order  so  long  as  the  volume  is  continuous. 

For  a  solid  of  revolution  with  the  2-axis  as  the  axis  of 
revolution,  the  formula  (1)  for  the  volume  becomes 


■// 


P2smdd6dp,  (2) 

since  the  limits  for  $  are  then  evidently  0  and  2  w.  The 
limits  for  p  and  0  are  then  the  same  as  are  used  in  getting 
the  area  of  the  plane  figure  revolved. 

Corollary.  —  dpe^V  =  p2  sin  0  dd  d<t>  dp  is  an  elementary 
rectangular  parallelopiped;  and  dp62V  =  27rp2  sin  Odd  dp  is  a 
circular  ring  with  rectangular  section. 

Example  1.  —  Find  the  volume  of  a  sphere  of  radius  a, 
using  polar  coordinates,  pole  at  end  of  a  diameter. 

By  formula  (1);  or  by  (2),  if  the  volume  is  considered  as 
generated  by  revolving  the  semicircle  about  the  initial  line, 
the  line  from  which  0  is  measured, 

I  p2  sin  d  dd  dp 

0     t/0 
x 

X 

r_  cos4  oy 

L  4     Jo 


167ra3    r      3.  .    - Aa 
-r —  /     cos3  0  sin  0(20 

O         t/0 

167ra3f 


war. 


324 


INTEGRAL  CALCULUS 


Example  2.  —  Find  the  volume  generated  by  revolving  the 
cardioid,  p  =  2  a  (1  —  cos  6)  about  the  initial  line. 


JrV      /*2a(l-cos0) 
/  p2  sin  6  dd  dp 

o    Jo 


167ra; 


j: 


(1  -  cos  0)3  sin  Odd 


64 


ira6. 


Example  3.  —  Find  the  volume  made  by  revolving  the 
lemniscate  p2  =  a2  cos  2  6  about  the  initial  line. 


*a  v  cos  2  0 


V  =  2tt  I      f  p2sinddddp 

Jo   Jo 

=  ^-  fT(cos20)*sin0d0  =  ^-  r(2cos20-l)tsin<9 
o    Jo  6    Jo 

V        2  V2  6/ 


ctt 


170.  Volumes  by  Double  Integration  —  Cylindrical  Co- 
ordinates. —  In    finding   the   volume    of    some    solids   the 

integration  is  performed  more 
readily  with  the  use  of  cylin- 
drical coordinates. 

In  this  system  of  coordi- 
nates the  position  of  a  point 
is  given  by  the  cylindrical  co- 
ordinates (r,  <f>,  z),  where  (r,  0) 
are  the  polar  coordinates  of 
the  projection  (x,  y,  0),  on  the 
XY  plane,  of  the  point  (x,  y,  z). 

It  is  evident  that  the  equa- 
tions of  transformation  from 
rectangular  to  cylindrical  co- 
ordinates are: 


- 

p 

J? 

A 

7 

f2 

\5t 

\T 

p> 

p' 

Q' 


r  cos  4>,     y  =  r  sin  <j>,     z  =  z; 


(i) 


VOLUMES  BY  DOUBLE  INTEGRATION  325 

and  those  from  cylindrical  to  rectangular, 
r  =  Vx2  +  y2,     </>  =  cos"1  -  =  sin"1  ^  =  tan"1  ^,     z  =  z.     (2) 

II  X 

To  derive  a  formula  for  volume  the  differential  element  of 
area  in  the  XY  plane  may  be  taken  as  the  rectangular  base 
of  an  elementary  right  prism  with  altitude  z,  the  base  of  the 
actual  prisms  into  which  the  solid  may  be  divided  being 
bounded  by  lines  two  only  of  which  are  right  lines,  the  other 
two  being  circular  arcs,  and  the  altitude  of  possibly  only  one 
edge  being  z,  since  the  surface  of  the  solid  may  be  curved, 
or  not  parallel  to  the  XY  plane,  even  when  plane. 
The  expression  for  the  volume  of  the  solid  is 


-// 


zr  d<t>  dr,  (1) 


where  z  must  be  expressed  in  terms  of  r  or  0  in  order  to  effect 
the  integration,  and  where  the  limits  are  to  be  such  as  will 
give  the  volume  sought. 

Corollary.  —  dr<p2V  =  zr  d<j>  dr  is  a  right  prism  with  rectangu- 
lar base.  The  double  integral  in  (1)  is  the  limit  of  the  sum  of 
the  elementary  solids  into  which  the  given  solid  is  conceived 
to  be  divided,  or  it  is  to  be  considered  simply  as  the  anti- 
differential  of  a  second  partial  differential,  when  the  differ- 
entials are  taken  as  finite  constants.  Either  way  of  regard- 
ing the  differentials  leads  to  the  same  result. 

Example  1 .  —  To  find  the  volume  of  a  sphere  of  radius  a. 
Taking  the  pole  at  the  center  of  the  sphere,  by  (1), 

7  =  2  /       faVa2-r2rd<t>dr 
Jq     Jo 


<*r 


d<j) 


3 


Example  2.  —  A  cylindrical  core  with  b  as  the  radius  of  a 
right  section  is  cut  from  a  sphere  of  radius  a.     Find  the 


326  INTEGRAL  CALCULUS 

volume  of  the  remaining  portion  of  the  sphere,  when  b  <  a 
and  the  axis  of  the  core  includes  a  diameter  of  the  sphere. 

7  =  2   r*  faVa2-r2rd<}>dr 

Jo     Jb 

=  2 £* d*[-  \{a2  -  r2)*]" 
=  f7r(a2-&2)*; 
/.      Vol.  of  core  =  y  (a3  -  (a2  -  62)*). 

Example  3.  —  Find  the  volume  in  first  octant  cut  from  a 
right  cylinder,  with  its  base  of  radius  ri  on  IF  plane  and  axis 

X         TJ         Z 

the  2-axis,  by  the  plane   -  +  j-  +  -  =  1 .     Here 

CL         0  C 

*  =  *(l-f-|)-«(l -£«»*- jjsin*). 

V  =  c  J      I    ( 1 cos  0  —  r  sin  4)  J  r  d$  dr 

rw     n3f  cos  <t>  ,  sin<A~|  .  2  Tt     n/l.lYl 

Example  4.  —  Find  the  volume  of  Ex.  8,  Exercise  XXXIX, 
by  using  formula  (1),  which  will  give  the  volume  sought 

more  easily  than  by  /    /    I  dx  dy  dz. 

IT 

P1*     fa  cost      , 

7  =  4/      /  Va2  -  r2rd<t>dr  =  §  (tt  -  J)  a3. 

Example  5.  —  Find  the  volume  of  the  segment  of  the  right 
cylinder  which  has  its  base  a  loop  of  the  lemniscate  r2  = 
a2  cos  2  0  in  the  XY  plane  and  its  upper  surface  a  plane 
which  intersects  the  XY  plane  in  the  ?/-axis  at  an  angle  of  45°. 


MASS  327 

Here  z  =  x  =  rcos<t>; 

cos  2  0 


Jo    Jo 
3  Jo 


r  cos  0r  d<t>  dr 
cos5  2  0  cos  <f>  d  <f> 


7T 

=  ^"-f  (l-2sin20)tCos0^ 

7T  V2  a3 
16 

171.  Mass.  Mean  Density.  —  As  stated  in  Art.  154, 
the  mass  of  a  body,  being  defined  as  the  product  of  density 
and  volume,  when  the  density  *  varies  continuously, 

m  =  limitT7A7=    fydV,  (1) 

which  becomes  m  =  j    J    j  ydxdydz,  (2) 

or  m  =   j    J   j  t  p2  sin  6  ddd(f>  dp,  (3) 

according  as  rectangular  or  polar  elements  of  volume  are 
used.  In  these  expressions  y  denotes  the  varying  density 
at  the  different  points  within  the  body.  The  mean  density 
of  the  body,  denoted  by  y,  is  given  by  the  equation 


fydV 


-      m 

y=y  =  —y-  '  (4) 

When  the  mass  is  considered  as  distributed  continuously 
over  a  surface,  the  element  of  volume  dV  is  replaced  by 
dA  =  dx  dy  or  pdd  dp;   and  when  the  mass  is  considered  as 

*  Density  will  now  be  denoted  by  7,  instead  of  p  used'in  Art.  154. 


328  INTEGRAL  CALCULUS 

distributed  along  a  line,  straight  or  curved,  dV  is  replaced 
by  ds,  the  element  of  length. 

When  the  integral  is  considered  as  an  anti-differential  the 
elements  are  expressed  directly  in  terms  of  the  finite  differen- 
tials; when,  however,  the  integral  is  considered  as  the  limit 
of  a  sum  or  sums,  the  elements  are  expressed  in  terms  of  the 
infinitesimal  increments,  the  differentials  appearing  under 
the  integral  sign. 

Example  1.  —  Find  the  mean  density  of  a  sphere  in  which 
the  density  varies  as  the  square  of  the  distance  from  the 
center. 

Here  the  distance  p  of  a  volume  element  determining  its 
density,  the  polar  element  should  be  used. 

Taking  the  density  at  a  distance  p  from  the  center  as 
kp2,  k  being  a  constant,  and  the  volume  element  as 
p2  sin  0  A0  A<£  Ap,  from  (4) , 

I      I       J    ftp4  sin  B  dO  d^>  dp 
—         Jo   Jo     Jo  o  7    9 

7  =  — ■ — = =  ■=  ka~. 

1 7ra3  5 

Again,  since  the  density  is  the  same  for  all  points  at  the  same 
distance  p  from  the  center,  taking  for  the  volume  element 
a  spherical  shell  of  thickness  Ap,  A  V  =  4  irp2  Ap,  whence 


7r  I    kp4  dp 


=  |fca2. 
o 


The  mean  density  is  thus  shown  to  be  three-fifths   the 
density  at  the  surface  of  the  sphere. 

[The  mean  density  of  the  earth  according  to  the  best 
determinations  is  very  nearly  5.527  times  that  of  water, 
while  the  average  density  of  rocks  at  or  near  the  surface 
is  only  about  two  and  a  half  times  that  of  water;  hence, 
the  mean  density  of  the  earth  is  about  twice  the  average 
density  at  the  surface.!    See  Corollary  at  end  of  Art.  190. 


mass  329 

Example  2.  —  Find  the  mass  and  mean  density  of  a  semi- 
circular plate  of  radius  a,  whose  density  varies  as  the  distance 
from  the  bounding  diameter.     Here  7  =  ky, 


m 


/  kydxdy  =  §/ca3; 

■aJo 


-  _  I  ka?  _4:ka 

\  TCL2         3  IT 


Or  m  =   I     I    kp2  sin  ddddp  =  §  ka3, 

Jo  Jo 

where  ky  —  kr  sin  6. 

By  a  single  integration,  the  element  of  area  being 


x  •  Ay  =  Va2  —  y2  &y, 
m  =  2  JkVa2-y2ydy  =  %ka\ 

Example  3.  —  Find  the  mean  density  of  a  straight  wire  of 
length  I,  the  density  of  which  varies  as  the  distance  from 
one  end. 

-f 

7  = 


ks  ds      7 , 
_  Id 

1         ~  2 ' 


Example  4.  —  Find  the  mass  and  mean  density  of  a  hemi- 
spherical solid,  radius  a,  the  density  varying  as  the  distance 
from  the  base. 


m 


fkzTX2  dz  =   I    kir  (a2  -z2)zdz  =  \  irka4; 
Jo  Jo 

lirka4      3 


Itto3       8 


ka. 


Here  the  element  of  volume  is  a  spherical  segment,  ttx2  A2  = 
-k  {a2  —  z2)  Az,  at  a  distance  z  from  the  base. 

Example  5.  —  Find  the  mean  density  of  a  right  circular 


330  INTEGRAL  CALCULUS 

cone  of  height  h,  in  which  the  density  varies  as  the  distance 
from  a  plane  through  the  vertex  perpendicular  to  the  axis. 

/a2  fh 
kz-wx2  dz      kir  r-  I    z?  dz      0 
h2J0  =  3  , , 

7~       iira2h       ~        iira2h  4 

Here  origin  is  taken  at  the  vertex  and  element  of  volume  is 

TTX2  Az  = 

a  being  radius  of  base;  y  =  kz. 


of 

TTX2  AZ  =  7T  t;  Z2. 

h2 


CHAPTER  VI. 
MOMENT  OF  INERTIA.     CENTER  OF  GRAVITY. 

172.  Moment  of  a  Force  about  an  Axis.  —  The  moment 
of  a  force  about  an  axis  perpendicular  to  its  line  of  direction 
is  the  product  of  the  magnitude  of  the  force  and  the  length 
of  the  perpendicular  from  the  axis  to  the  line  of  action  of 
the  force.  The  moment  is  the  measure  of  the  tendency  of 
the  force  to  produce  rotation  about  the  axis. 

The  moment  of  a  force  about  a  point  is  identical  with  the 
moment  about  an  axis  through  the  point,  perpendicular  to 
the  plane  containing  the  point  and  the  line  of  action  of  the 
force. 

173.  First  Moments.  —  Let  a  line,  surface,  or  solid  be 
divided  into  elements;  let  each  element  (As,  AA  or  AT)  be 
multiplied  by  the  distance  of  a  chosen  point  within  the 
element  from  a  reference  line  or  plane. 

The  limit  of  the  sum  of  these  products  as  the  elements  are 
taken  smaller  and  smaller  is  called  the  first  moment  of  the 
line,  surface,  or  solid. 

For  the  first  moment  Mx  of  a  plane  curve  about  the  x-axis, 

Mx  =  \im  Vy\s  =    I  yds;  (1) 

As=0  **  J 

and  for  the  first  moment  of  a  plane  area  about  the  same  axis, 

M,  =  lim  T?/AA  =    fydA.  (2) 

The  first  moment  of  a  solid  with  respect  to  one  of  the  co- 
ordinate planes,  say  the  XY  plane,  is  given  by  the  equation, 


Mxy  =  lim 

331 


%z±Y  =fzdV.  (3) 


332  INTEGRAL  CALCULUS 

For  A  A  and  AF  appropriate  expressions  for  the  area  and 
volume  elements  are  to  be  used  and  the  values  corresponding 
for  dA  and  dV  substituted,  in  order  to  effect  the  integrations. 
The  elements  may  be  so  taken  that  a  single  integration  is 
sufficient,  but  double  or  triple  integration  will  in  general  be 
required. 

174/  Center  of  Gravity  of  a  Body.  —  Let  a  given  mass  be 
referred  to  a  system  of  rectangular  coordinates,  and  let  Mxy, 
Myz,  Mxz,  denote  the  first  moments  with  respect  to  the  three 
coordinate  planes. 

The  first  moments  of  the  mass  of  a  solid  are  derived  from 
those  of  the  geometrical  solid  by  the  introduction  of  a 
density  factor. 

There  will  be  a  point  G  (x,  y,  i),  given  by  the  equations, 

JyxdV         fyydV  fyzdV 

m=  f  ydV;  x  =  — ,  y  =  -» -,  z  =  -^ -,  (4) 

I  ydV  J  ydV  J  ydV 

in  which  the  letter  y  denotes  the  density. 

The  point  G  thus  determined  is  the  centroid  of  the  mass. 
It  is  also  the  center  of  gravity  of  the  weight  W.  Since  W  = 
mg,  the  masses  of  particles  of  a  body  are  directly  proportional 
to  their/Weights;  hence,  the  center  of  gravity  is  the  same  as 
the  center  of  mass.  The  force  of  gravity  acting  on  any  mass 
is  an  example  of  a  force  distributed  through  a  volume.  If 
w  denote  the  weight  per  unit  of  volume,  at  any  point  in  a 
given  mass,  W  its  entire  weight,  and  x,  y,  z,  the  coordinates 
of  the  center  of  gravity,  the  point  where  the  resultant  force 
exerted  by  gravity  would  act;  then,  from  (4)  or  (3), 

//  wxdV           j  ivydV           I  wzdV 
wdV;  x=^ ,  y=4- ,  z  =  ^ .    (5) 

JwdV  wdV  IwdV 

If  in  (4)  and  in  (5),  y  and  w  are  constant,  that  is,  if  the  mass 


CENTER  OF  GRAVITY  OF  A  PLANE  SURFACE      333 

is  homogeneous,  they  may  be  taken  from  under  the  integral 
sign  and  canceled;  whence, 


JxdV  fydV  CzdV 


m=yV,W  =  wV;  x= — y — ,y  = — —  ,  z  =  ,    (6) 

the  coordinates  of  the  centroid  of  a  volume,  or  of  a  homo- 
geneous body. 

The  quantities   /  xdY  =  zV,    j  y  dV  =  yV,    j  zdV  =W, 

equal  the  moments  of  the  volume  with  respect  to  the  YZ,  XZ, 
and  XY  planes,  respectively. 

175.  Center  of  Gravity  of  a  Plane  Surface.  —  If  in  the 
formulas  (6)  for  the  centroid  of  a  volume  dV  is  replaced  by 
t  dA  and  i  taken  equal  to  zero,  then 

fxtdA       CxdA  fytdA      fydA 

V  =  tA,x  =  +p =  ^ ,  y=^- =  ^- ,    (1) 

J  tdA        I  dA  t  dA         /  dA 

where  the  point  (x,  y)  is  in  the  XY  plane,  and  in  winch  dA 
is  the  area  of  an  element  of  the  surface  of  a  thin  plate  of 
uniform  thickness  and  material,  making  t  dA  =  dV.  If  A 
be  the  area  of  the  middle  layer,  it  is  evident  that 


xA  =    I  xdA     and     yA  =    f  ydAi 


(2) 


which  are  called  the  moments  of  the  area  A  with  respect  to 
the  y-axis  and  the  x-axis.  By  the  center  of  gravity  of  a 
plane  surface  is  meant  that  point  which  is  the  center  of 
gravity  of  a  thin  plate  of  uniform  thickness  and  material 
whose  middle  layer  is  the  surface  given.  The  formulas  for 
its  coordinates  are,  therefore,  those  given  in  (1), 

j  xdA  I  ydA 

*  =  — a — >  y  =  — a — 


334  INTEGRAL  CALCULUS 

It  is  evident  that  the  moment  of  an  area  about  an  axis 
through  its  center  of  gravity  will  be  equal  to  zero. 

176.  Center  of  Gravity  of  any  Surface.  —  The  formulas 
(6),  Art.  174,  become  for  any  surface, 

I  xdA  j  ydA  j  zdA 

5=-X-,  v  =  ^r~ '   i  =  ^—       (1) 

By  the  same  method  as  in  Art.  174,  it  can  be  shown  that  the 
coordinates  of  the  center  of  gravity  of  any  surface,  plane  or 
curved,  are  given  by  the  equations  (1). 

177.  Center  of  Gravity  of  a  Line.  —  If  in  the  formulas 
(5),  Art.  174,  for  the  coordinates  of  the  center  of  gravity  of 
a  body  of  weight  W,  w  dV  is  replaced  by  w  ds,  and  2  =  0; 
then, 

/  wxds       j  xds               j  wyds        lyds 
i=— i =  '   ,     y  =  — =  — ,       (1) 

j  wds  I  ds  I  wds  I  ds 

where  the  point  (x,  y)  is  in  the  XY  plane,  and  in  which  w  ds 
is  the  weight  of  an  elementary  length  of  a  slender  rod  of 
uniform  section  and  material  whose  weight  per  unit  of 
length  is  equal  to  w.  If  s  be  the  length  of  the  center  line  of 
the  rod,  it  is  evident  that 


=    I  xds    and     ys  =    j 


xs  =   j  xds    and     ys  =    I  y ds,  (2) 

which  are  called  the  moments  of  the  line  with  respect  to  the 
2/-axis  and  the  z-axis.  The  rod  may  be  straight,  in  which 
case  the  center  of  gravity  will  be  at  the  middle  point  of  its 
center  line.  If  the  center  line  of  the  rod  is  a  plane  curve  in 
the  plane  XY,  the  coordinates  of  the  center  of  gravity  are 
given  by  (1).  By  the  expression,  center  of  gravity  of  a  line, 
is  meant  the  point  which  is  the  center  of  gravity  of  a  slender 
rod  of  uniform  section  and  material,  of  which  the  given  line 


CENTER  OF  GRAVITY  OF  A  SYSTEM  OF  BODIES        335 

is  the  center  line.  The  coordinates  of  a  line  are,  therefore, 
those  given  in  (1).  It  is  evident  that  the  moment  of  a  line 
about  an  axis  through  its  center  of  gravity  will  be  equal 
to  zero. 

If  the  center  line  of  the  rod,  or  any  given  line,  is  not  a  plane 
curve,  from  (5)  as  before,  the  equations  are 

j  xds  I  yds  I  zds 

x  =  -~ '    y  =  ~p '    5  =  -p 

I  ds  j  ds  j  ds 

The  moments  of  the  line  with  respect  to  the  YZ,  XZ,  and 
XY  planes,  respectively,  will  be 


xs 


=   f  xds,     ys  =    I  yds,    zs  =   I  z ds. 


178.   Center  of  Gravity  of  a  System  of  Bodies.  —  If, 

instead  of  a  single  body,  there  is  a  system  of  bodies  whose 
volumes  are  Vi,  V2,  V3,  .  .  .  Vn,  the  coordinates  of  their 
centers  of  gravity  being,  respectively,  (xh  yh  zi),  etc.;  and, 
if  (x0j  2/0,  io)  denote  the  coordinates  of  the  center  of  gravity 
of  the  system  and  Vo  its  total  volume,  then 

V0  =  Vx  +  V2  H-  Vz  +  •  •  •  +  Vn', 

Myg  =   VlX!  +  V2X2  +     •     •     •     VnXn  =    J)  Vx=    V&). 

Similarly, 

Mxz  =  J)  Vy  -=  V0yo,     and    Mxy  =^Vz=  V0z0. 

Hence, 

-_%Vx      _  _%Vy      _  _^V~z 

Xo~~v^>   yo~~v^>  Zo~~v^' 

where  the  numerators  are  the  sum  of  the  moments  of  the 
system  with  the  respective  coordinate  plane,  the  equalities 
following  from  equations  (6)  of  Art.  174.  Similar  equations 
hold  for  weights  or  masses  upon  substituting  W  or  m  for  V, 
and  for  any  group  of  surfaces  by  substituting  A,  where  A0 


336  INTEGRAL  CALCULUS 

is  the  sum  of  the  areas  of  the  several  surfaces.     If  the  sur- 
faces are  plane, 


_  ^Ax      _  _*%Ay 


The  last  equations  are  useful  in  getting  the  center  of 
gravity  of  plane  figures  composed  of  parts,  the  centers  of 
gravity  of  which  are  known  or  easily  found. 

Similar  equations  hold  for  a  system  of  lines;  s0  being  the 
sum  of  the  lines, 

Xs^      -      Xs2/      _      Xs* 

£o  =  — — ,     2/=      — >     z  =  — — i 

So  So  So 

if  the  lines  are  not  all  in  one  plane;  and  zo  =  0  when  they 
are  in  one  plane,  taken  as  the  plane  of  XY. 

179.  The  Theorems  of  Pappus  and  Guldin  —  First 
Theorem.  —  An  arc  of  a  plane  curve  revolving  about  an  axis  in 
the  plane  of  the  curve,  but  not  intersecting  it,  generates  a  surface 
of  revolution,  the  area  of  which  equals  the  product  of  the  length 
of  the  revolving  arc  and  the  length  of  the  path  described  by  its 
center  of  gravity. 

Second  Theorem.  —  A  plane  area,  bounded  by  a  closed 
curve,  revolving  about  an  axis  in  its  plane  but  outside  the  area, 
generates  a  solid,  the  volume  of  which  equals  the  product  of  the 
revolving  area  and  the  distance  traveled  by  its  center  of  gravity. 

To  prove  the  first  theorem;  let  the  z-axis  be  the  axis  of 
revolution,  then  the  surface  generated  by  the  revolution  of 
the  curve  about  the  z-axis  is 


rfV 


ds.     (1)  Art.  160. 
From  (2),  Art.  177,  for  a  plane  curve, 

ys  =  f  y  ds. 
Hence  S  =  2irys.  (1) 


THE  THEOREMS  OF  PAPPUS  AND  GULDIN       337 

It  is  evident  that,  if  only  part  of  a  revolution  is  made,  the 
area  of  the  surface  generated  will  be  given  by 

Si  =  6ys,  (2) 

where  8  denotes  the  angle  in  radians  through  which  the  plane 
containing  the  curve  is  turned.  It  is  to  be  noted  that  the 
theorem  and  proof  include  the  case  of  a  segment  of  a  straight 
line  revolving  about  any  axis. 

To  prove  the  second  theorem;  let  the  z-axis  be  the  axis 
of  revolution,  as  before;  then,  denoting  by  A  A  an  element 
of  the  plane  area,  the  volume  generated  by  a  complete 
revolution  of  the  area  is 

V  =  lim  y.2iry\A  =  2w  fydA. 

Now  JydA  =  yA;     (2)  Art,  175; 

hence,  V  =  2wyA.  (3) 

It  is  evident  that  if  only  part  of  a  revolution  is  made,  the 
angle  turned  through  by  the  plane  of  the  area  being  0 
radians,  the  volume  generated  will  be  given  by 

Vi  =  By  A-  (4) 

Example  1.  —  Find  the  center  of  gravity  of  a  semicircle  of 
radius  a.     Find  it  for  the  semicircular  arc  also. 

Taking  the  diameter  along  the  ?/-axis,  the  length  of  the 
path  described  by  the  center  of  gravity  as  the  semicircular 
area  is  revolved  about  the  y-axis  is  2  irx.  The  semicircle  by 
its  revolution  describes  a  sphere  whose  volume  is  $  ira3; 
hence,  by  the  second  theorem  of  Pappus, 


\  irx  •  \  wa2  =  $  7ro3, 


_      4a 
&  =  =—■ 

0  7T 


338 


INTEGRAL  CALCULUS 


Also  the  arc  describes  the  surface  of  a  sphere,  47ra2;  hence, 
by  the  first  theorem  of  Pappus, 

2  ttx  -  ira  =  4  ira2 ; 

_      2a 
.*.     x  = 


Example  2.  —  Find  the  center  of  gravity  of  the  semi- 

ellipse,  -2  +  ^  =  1. 

Taking  the  z-axis  as  the  axis  of  symmetry  and  applying 

the  second  theorem  of  Pappus 
as  in  Ex.  1,  it  is  found  that 

x  =  5— ,  also ;  hence,  as  y  =  0, 

the  centers  of  gravity  of  the 
two  areas  are  identical. 

Example    3.  —  Find    the 
center    of    gravity    of    the 
^4     x     quadrant  of  the  ellipse. 

Let  AOB  be  the  quadrant 
of  the  ellipse,  and  the  element  of  area  a  narrow  strip  parallel 
to  !/-axis. 


dA  =  ydx  =  -  Va2 
*  a 


x2dx. 


x  = 


/xdA      -  I    x  Va2  —  x2  dx 
a  Jo 

A  j  Tab 

a\_  Jo  _  4 a     /The  same  abscissa\ 

i  7ra6  ~  3  7r     \as  for  semi-ellipse.  / 


_46 
Similarly,  it  is  found  that  y  =  ^—  • 

Corollary.  —  For  the  circle,    x2  +  y1 


2    -      -      4a 
a2;x  =  y  =  ^. 


EXAMPLES  OF  CENTER  OF  GRAVITY 


339 


Example  4.  —  Find  the  center  of  gravity  of  a  circular  arc. 
Let  AB  be  a  circular  arc,  whose  radius  is  a  and  whose  center 
is  at  origin  0.     Let  0i  and  02  be 
the  angles  with  x-axis  made  by 
the  radii  OB  and  OA  to  the  ends 
of  the  arc.     From 


.  / 


/«• 


xds  I  yds 

and  y  =  * 


I 


ds 


using   polar   coordinates,    x  =  a    o 
cos  6,  y  =  a  sin  6,  ds  =  a  dd; 


\J 


a2 cos  0  dd 


•     T 

a  sin  0 

/    a  dO  e\ 

Je2  Je2 


a  (sin  di  —  sin  02) 

di  —  &2 


and 


2/  = 


re,                                       10! 
I     a2  sin  Odd       —  a  cos  0  ,       „  „  N 

Jg2 J^  _  a  (cos  02  —  cos  0i) 

a  dd  d 


01  —  02 


When  02  =  0,  the  equations  reduce  to 
a  sin  0i 


x  = 


a  (1  —  cos0i) 

»■ — * — 


Corollary.  —  When  0i  =  0i,  02  =  —  0i;   x 


When 


0X  =  90°,    02  =  0;    x  = 


a  sin  0i 


2a 


y  =  o. 


2_a_ 

7T 


Example  5.  —  Find  the  center  of  gravity  of  a  triangle. 
Let  the  triangle  ABC  have  base  b  and  altitude  A,  and  let  the 
x-axis  be  through  the  vertex  parallel  to  the  base,  and  the 
2/-axis  positive  downwards.     Take  dA  =  L  dy,  where  L  is 


340 


INTEGRAL  CALCULUS 


the  length  of  an  elementary  strip,  parallel  to  the  base  and  at 
a  distance  y  from  z-axis. 


dA  =  ^ydy; 


hence 


L  :y  =  b  :h; 


-i- 


A  ib) 

Similarly,  by  taking  strips  parallel  to  h  or  the  2/-axis, 

I  xdA  _t  J    x2dx 


x  = 


hbh 


b. 


A 

Y" 

0 

X 

1            /I 
1         /l 

i 
h 

i 
i 

Y 

-i»y 

////M/M 

D 

/    i 

/! 
1% 

/     ! 

m 

C 

Y 

t* 

-b > 

Hence  the  perpendicular  distance  of  the  center  of  gravity 
from  the  base  will  be  \  h,  and,  since  the  center  of  gravity  of 
each  elementary  strip  will  be  its  middle  point,  the  center  of 
gravity  of  the  triangle  will  be  on  the  median  line  Am,  at 
one-third  the  distance  from  m  to  A.  Similarly,  it  is  on  the 
median  line  from  B;  hence  it  is  at  the  intersection  of  the 
medians. 

Example  6.  —  Find  the  center  of  gravity  of  a  semicircular 
plate  of  radius  a,  whose  density  varies  as  the  distance  from 
the  center. 

Here,  the  density  being  determined  by  the  distance  from 
the  center,  the  polar  element  is  used. 


EXAMPLES  OF  CENTER  OF  GRAVITY  341 

Let  the  density  y  =  kp,  and  the  z-axis  as  in  Ex.  1 ;  then 


x  = 


JyxdA      fSkr3cosededp 


f 


ydA 


J      jkp2dedP 
4j    . 

4f. 


cos  6  dd 


3a 


and     y  =  0. 


de 


Example  7.  —  Find  the  center 
of  gravity  of  the  volume  cut  from 
a  right  cylinder,  the  radius  of 
whose  base  is  a,  by  the  planes 
z  =  0  and  z  =  mx,  the  volume 
above  the  plane  of  XY  alone  con- 
sidered.    Here 

h 
x 

x  dx  dy  dz 


nvtf-^      pa 
/ 

a  Jo 


vV 


__  Tra^h'S  _  Ta3h/S 
X~       V  ia2h 

2  „2 


x2  dx 

3_ 

16 


TQ, 


7=f  a2h  from  Ex.  2,  Exercise  XXXIX. 


I  zdxdy  dz 

0    ./-Vfli-jiJo 

=  -=   /    x2  Va2  —  x2  dx  = 
a2  Jo 


iraW 
16 


_      Tra2h2/1Q      Tra2h2/\(S 

Z  =  77 =         o      o,     - 


32 


irh;    and     y  =  0. 


342  INTEGRAL  CALCULUS 

EXERCISE  XL. 

Find  the  coordinates  of  the  center  of  gravity: 

1.  Area  of  the  parabolic  half  segment,  y-  =  4  ax,  x  =  0  to  x  =  a. 

Ans.  (f  a,  \a). 

2.  The  area  under  one  arch  of  the  cycloid,  x  =  a  (0  —  sin  0), 
y  =  a  (1  —  cos0).  Ans.  (wa,  $  a). 

3.  The  area  under  the  catenary,  y  =  -  ^e«  +  e    aJ,x  =  0tox  =  a. 

AnS'   \e  +  V         8(e-e-i)      ) 

4.  A  semicircular  plate  of  radius  a,  the  density  varying  as  the 
distance  from  the  bounding  diameter.  Ans.  (T%7ra,  0). 

5.  A  homogeneous  right  circular  cone,  radius  of  base,  a,  and  altitude 
h.     The  axis  on  the  z-axis.  Ans.  x  =  %h,y  =  z  =  0. 

6.  A  homogeneous  paraboloid  of  revolution  from  the  origin  to  x  =  h. 

Ans.  x  =  f  h,  y  =  0. 

7.  A  hemisphere  whose  density  varies  as  the  distance  from  the  base 
whose  radius  is  a.  Ans.  (0,  0,  ^  a.) 

8.  The  eighth  part  of  a  sphere  in  the  first  octant,  the  density  of  the 
mass  varying  as  the  distance  from  the  pole  or  origin  at  the  center. 

Ans.  x  =  y  =  |  a. 

9.  The  circular  sector  subtending  the  angle  6h  radius  a. 

t  \    t?     n       a  (  \   a        /2a(sin0i)     2a(l  —  cos0i)\ 

(a)  Fore,  =  0,  (a)  Am.  (3—^—,  3 5-  -2} 

(b)  For  0,  =  90°,  the  quadrant.  (6)  Ans.  x  =  y  =  P^- 

O  TV 

4  a 

(c)  For  0i  =  180°,  the  semicircle.  (c)  Ans.  x  =  5— ,  y  =  0. 

10.  Find  the  center  of  gravity  of  a  T-section.  Using  the  method  of 
Art.  178,  with  the  dimensions  on  figure,  taking  the  moments  of  the  three 
rectangles  composing  the  section, 


%Ay 


4X7^  +  6X4  +  8X}  _58_  29 


y  =  sr  =  —  4  +  6  +  8 — -  -  is  -  9  -  3-22|h19- 

x0  =  0,  as  the  center  of  gravity  will  be  on  the  axis  of  symmetry.  Here 
advantage  is  taken  of  the  knowledge  that  the  center  of  gravity  of  each 
of  the  rectangles  is  at  the  center  of  the  rectangle. 

Note.  —  This  is  the  method  of  finding  the  centers  of  gravity  of  the 
various  shapes,  or  built-up  sections,  used  in  constructions. 


EXAMPLES  OF  CENTER  OF  GRAVITY 


343 


v\ 


<7|- 


y? 


< 8" 

11.  Find  the  center  of  gravity  of  the  trapezoid  OAao.  Let  the  upper 
base  be  b  and  the  lower  base  B,  the  altitude  h.  Divide  the  trapezoid 
into  two  triangles  by  diagonal  Oa,  then  the  distances  of  the  centers  of 
gravity  from  OA"  are  |  h  and  f  h  for  the  triangles,  respectively.     Then 


W 


A0 


X    3 


(B  +  b) 


h/B  +  2b\ 
3\B+b  J 


The  center  of  gravity  of  each  strip  of  area  parallel  to  the  bases  will  be 
its  middle  point,  hence  the  center  of  gravity  of  the  whole  area  is  on  the 
median  line  mhi  of  the  trapezoid.  Graphically,  the  center  of  gravity 
is  located  by  again  dividing  the  trapezoid  into  two  other  triangles  by 
the  diagonal  oA,  and  drawing  lines  connecting  each  pair  of  centers  of 
gravity;  the  intersection  of  these  connecting  lines  is  the  center  of 
gravity  G  of  the  trapezoid. 


344  INTEGRAL  CALCULUS 

12.  From  Art.  178,  show  that  the  center  of  gravity  of  two  volumes, 
masses,  areas,  or  lines,  lies  on  the  line  joining  their  separate  centers  of 
gravity  and  divides  that  line  into  segments  inversely  proportional  to 
the  two  magnitudes. 

In  Ex.  11,  the  point  G,  the  center  of  gravity  of  the  trapezoid  divides 
the  line  GiG2  connecting  the  centers  of  gravity  of  two  triangles,  inversely 
as  the  areas  of  the  triangles,  and  it  divides  the  line  G'G"  in  the  same 
way. 

In  this  way  the  common  center  of  gravity  of  the  Earth  and  the  Sun 
is  found.  Taking  the  distance  from  the  Earth  to  the  Sun  as  92,400,000 
miles,  and  the  mass  of  the  Sun  as  327,000  times  that  of  the  Earth, 
makes  the  distance  of  the  common  center  of  gravity  from  the  center  of 

the  Sun      '       '         miles,  or  only  about  280  miles;  so  that  the  Sun  is 

considered  practically  at  rest  relative  to  the  Earth. 

180.  Second  Moments  —  Moment  of  Inertia.  —  The 
term  moment  of  inertia  is  applied  to  a  number  of  expressions 
which  are  second  moments  of  lines,  of  areas,  or  of  solids. 

Let  each  of  the  elements  of  length,  of  area,  or  of  volume 
(As,  AA,  or  AF),  into  which  a  line,  a  surface,  or  a  solid  may 
be  supposed  to  be  divided,  be  multiplied  by  the  square  of  the 
distance  of  some  chosen  point  in  the  element  from  a  reference 
line  or  plane. 

The  limit  of  the  sum  of  these  products  as  the  elements  are 
taken  smaller  and  smaller  is  called  the  secorid  moment  of  the 
given  line,  surface,  or  solid  with  respect  to  the  reference  line 
or  plane. 

Formulas  for  second  moments  are  derived  from  those  for 
first  moments  by  squaring  the  distance  factor.  Denoting 
the  second  moment  by  /,  the  general  symbol  for  moment  of 
inertia,  the  following  formulas  correspond  to  (1),  (2),  (3), 
Art.  173. 

For  a  plane  curve,      Ix  =   I  y2ds.  (1) 

For  a  plane  area,        Ix  =   f  y2dA.  (2) 

For  a  volume,  Ixy=fz2dV.  (3) 


POLAR  MOMENT  OF  INERTIA  345 

As  applied  to  an  area,  the  moment  of  inertia  is  a  numerical 
quantity  entering  into  a  large  number  of  engineering  com- 
putations and  takes  its  name  from  the  analogy  between  the 
mathematical  expression  for  it  and  that  for  the  moment  of 
inertia  of  a  mass  or  solid.  It  is  evident  from  the  form  of  the 
expression  that  the  moment  of  inertia  is  always  a  positive 
quantity,  being  unlike  the  first  moment  in  that  respect.  In 
distinction  from  the  moment  of  inertia,  the  first  moment  is 
sometimes  termed  the  statical  moment. 

181.  Radius  of  Gyration.  —  The  radius  of  gyration  of  a 
solid  is  the  distance  from  the  reference  line,  called  the  in- 
ertia axis,  to  that  point  in  the  solid  at  which,  if  its  entire 
mass  could  be  concentrated,  its  moment  of  inertia  would  be 
unchanged.  Thus,  if  m,  the  entire  mass  of  a  body,  be  con- 
sidered as  concentrated  at  a  point,  and  k  denote  the  distance 
from  the  inertia  axis  to  that  point,  the  expression  for  the 

moment  of  inertia,    /  r2  dm,  will  be  equal  to  k2m.     Therefore, 

I  =  k2m  and  k  =  y  — ,  k  being  the  radius  of  gyration  of 

the  solid  of  mass  m. 

In  the  case  of  the  plane  area,  by  analogy, 

Ix  =   Cx2dA  =  k2A,    and     k  =  y^-, 

where  A  is  the  entire  area  and  k  is  its  radius  of  gyration  with 
respect  to  the  a>axis. 

The  radius  of  gyration  of  both  the  solid  mass  and  the 
plane  area  will  evidently  be  expressed  in  linear  units. 

182.  Polar  Moment  of  Inertia.  —  The  polar  moment  of 
inertia  of  a  plane  area  is  the  moment  of  inertia  about  an  axis 
perpendicular  to  its  plane.  The  moments  of  inertia  about 
any  two  rectangular  axes  in  the  plane  of  the  area  are  called 
rectangular  moments  of  inertia  when  they  are  mentioned 
in  connection  with  the  polar  moment.     It  is  evident  that 


346 


INTEGRAL  CALCULUS 


for  an  area  in  the  plane  of  XY,  the  moment  of  inertia  about 
the  2-axis  is  a  polar  moment  of  inertia,  and  that  it  is  equal 
to  the  sum  of  the  two  rectangular  moments. 


Y 

x__mp 

~o 


Thus,  let  the  point  P  (x,  y,  0)  be  in  the  element  of  area, 

then,  L=    fr2dA 

is  the  polar  moment  of  inertia  of  the  area  about  the  z-axis  or 
with  respect  to  the  point  0  in  its  plane.     But  since 


r2  =  x1  +  y2,    I 


z  =  f  (y2  +  & 


)dA> 


hence,  Iz  =  Ix  +  Iv. 

The  symbol  for  the  polar  moment  of  inertia  is,  in  general, 
the  letter  J. 

183.  Moments  of  Inertia  about  Parallel  Axes.  — 
Theorem.  —  The  moment  of  inertia  of  an  area  about  an  axis 
in  its  plane,  not  passing  through  its  center  of  gravity,  is  equal 
to  its  moment  of  inertia  about  a  parallel  axis,  passing  through 
its  center  of  gravity,  increased  by  the  product  of  the  ana  and 
the  square  of  the  distance  between  the  two  axes. 


MOMENTS  OF  INERTIA  ABOUT  PARALLEL  AXES     347 

Let  the  origin  be  at  Og,  the  center  of  gravity  of  the  area, 
and  take  the  y-axis  parallel  to  the  inertia  axis  through  0  in 
the  plane.     Let  the  point  P  (x,  y)  be  in  the  element  of  area, 


<^a~> 

\° 

o9 

then  its  coordinates  with  respect  to  the  axis  through  0  are 
(x  +  a,  y).  The  moment  of  inertia  IG=  j  x2  dA,  and  that 
about  the  a:ds  through  0  is 

7=   f(x  +  a)2dA 
=  J(x2  +  a2  +  2ax)dA 
=   fx2dA+   Ca2dA+  ^2axdA 
=  IG  +  Aa2  +  2a  CxdA.  (1) 

The  quantity  I  x  dA  must  be  equal  to  zero,  since  the  y-zxis 

passes  through  the  center  of  gravity  (Art.  175).     Therefore, 

I  =  IG  +  Aa2.  (2) 

It  is  evident  that  equation  (1)  gives  the  relation  between 
the  moments  of  inertia  with  respect  to  any  two  parallel  axes 
in  the  plane  of  the  area,  OgY  being  replaced  by  an  axis 
through  any  point  in  the  plane. 

In  the  same  way  it  can  be  proved  that  the  polar  moment 
of  inertia  of  the  area,  with  respect  to  any  point  0,  is  equal 
to  its  polar  moment  of  inertia  with  respect  to  its  center  of 


348  INTEGRAL  CALCULUS 

gravity  plus  the  product  of  the  area  and  the  square  of  the 
distance  between  the  point  and  the  center  of  gravity. 

Corollary  1.  —  It  follows  from  (2),  that,  of  all  parallel  axes, 
the  axis  through  the  center  of  gravity,  called  the  gravity 
axis,  has  the  least  moment  of  inertia. 

Corollary  2.  —  When  the  inertia  axis  is  a  gravity  axis,  the 
radius  of  gyration,  then  called  the  principal  radius  of  gyra- 
tion, is  the  least  radius  for  parallel  axes;  from  (2), 

Ak2  =  AkG2  +  Aa2,     ,\     k2  =  kG2  +  a2,  (3) 

where  kg  is  the  principal  radius  of  gyration  and  a  is  the 
distance  between  it  and  a  parallel  axis. 

184.  Product  of  Inertia  of  a  Plane  Area.  —  The  product 
of  inertia  of  a  plane  area  is  a  numerical  quantity  which  is  of 
value  only  as  it  is  found  to  enter  into  the  determination  of 
the  relations  between  moments  of  inertia  with  respect  to 
different  axes.  The  product  of  inertia  of  an  area  A  with 
respect  to  the  axes  of  x  and  y  is  a  second  moment, 


/ 


xydA 


and  may  be  defined  as  the  limit  of  the  sum  of  the  products 
of  the  elementary  areas  and  the  product  of  their  distances 
from  the  two  coordinate  axes.  Unlike  the  moment  of 
inertia,  the  product  of  inertia  may  evident^  be  either  posi- 
tive or  negative,  depending  upon  its  distribution  in  the 
different  quadrants;  and  the  area  may  be  so  located  that 
its  product  of  inertia  will  be  zero. 

The  axes  may  be  so  chosen  as  to  make  the  product  of 
inertia  of  an  area  zero,  and  such  axes  are  called  principal 
axes,  the  corresponding  moments  of  inertia  being  called 
principal  moments  of  inertia.  It  can  be  shown  that  for  any 
point  of  an  area  (or  body)  there  exists  a  pair  of  rectangular 

axes  for  which  I  xy  dA  =  0,  and  that  the  moment  of  inertia 

is  a  maximum  when  taken  with  respect  to  one  of  the  principal 


LEAST   MOMENT  OF  INERTIA  349 

axes,  and  a  minimum  when  taken  with  respect  to  the  other. 
The  relation  between  the  polar  moment  and  the  two  rec- 
tangular moments  (Art.  182)  shows  that  if  one  of  the  two  is 
a  maximum  the  other  is  a  minimum,  and  vice  versa. 

185.  Least  Moment  of  Inertia.  —  In  designing  a  column 
an  engineer  needs  to  know  the  least  radius  of  gyration,  and 
consequently  the  least  moment  of  inertia,  of  the  cross  section, 
since  the  resistance  to  bending  is  least  about  that  axis  which 
has  the  least  moment  of  inertia.  It  has  been  stated  that 
the  moment  of  inertia  of  an  area  is  least  about  a  principal 
axis  through  the  center  of  gravity,  and  it  is  necessary,  there- 
fore, to  determine  those  principal  axes  which  pass  through 
the  center  of  gravity. 

In  many  cases  the  position  of  the  principal  axes  are  known 

at  once,  for  all  axes  of  symmetry  are  principal  axes,   /  xy  dA 

being  equal  to  zero  for  such  axes,  since  for  every  point  (x,  y) 
there  is  another  (x,  —y). 

For  example,  any  diameter  of  a  circular  area,  the  axis  of 
a  parabola,  either  axis  of  the  ellipse  or  of  the  hyperbola  is 
a  principal  axis.  For  a  rectangle  it  is  obvious  that  the  lines 
through  the  center  parallel  to  the  sides  are  principal  axes, 
but  the  diagonal  of  a  rectangular  plate  is  not  a  principal 
axis  at  its  middle  point.  The  gravity  axes  parallel  and 
perpendicular  to  the  web  of  the  cross  section  of  an  /-beam, 
channel,  or  T-beam  are  principal  axes. 

When  the  section  is  unsymmetrical,   it  is  necessary  to 

evaluate  the  integral,    I  xy  dA,  in  determining  the  principal 

axes. 

Remark.  —  A  full  treatment  of  the  subject  of  moment  of 
inertia  and  product  of  inertia  is  beyond  the  scope  of  tins 
book.  What  has  been  given  has  been  confined  for  the  most 
part  to  areas,  as  that  part  of  the  subject  has  more  immediate 
application  in  engineering. 


350 


INTEGRAL  CALCULUS 


186.  Deduction  of  Formulas  for  Moment  of  Inertia. 

1.   Rectangle  of  base  b  and  altitude  h: 

h 

Ix  =  ftfdA  =  f2by*dy  =  ^bh*. 

Gar-i 


la'v  =  h  +  Aa2  =  j2  W  +  hh 


P3- 


fx2dA  =    F  hx2dx  =  &¥h. 

~2 


Again, 


iv  =    /    by2  dy 

t/0 


=  \bh\ 
h  =  tV&4>    J  =  lbA;  for  square. 

Y 


A 

^9 

X'—- 
A' 

_j 

5' 

1 

(1) 

(2) 
(3) 


J  =  Ix  +  Iv  =  ~bW  +  ^Vh  =  b^  +  W).         (4) 


(2') 


2.    Triangle  about  the  axes: 

(a)    Through  the  apex  parallel  to  the  base. 

(6)   Through  the  center  of  gravity  parallel  to  the  base. 

(c)   Through  the  base. 


DEDUCTION  OF  FORMULAS 
Let  the  base  be  6  and  the  altitude  h. 

(a)  Ix  =  ftfdA=£\y>dy  =  h-^- 

(b)  I0  =  Ix-Aa*  =  T-Y(^-)  = 


bh? 
36' 


fM        t    .   a  2     bh*bh  /W\      bh* 


3.   Circle: 

(a)    Polar  moment  of  inertia,  axis  through  center. 
(6)    Moment  of  inertia  about  a  diameter. 


351 

(1) 

(2) 
(3) 


(a)    J  =   fr*  dA  =  Jr2  to3  dp  =  ~ 


Trrf4 
32 


(1) 


352                            INTEGRAL  CALCULUS 

For  a  sector  with  angle 

e, 

J 

-£•*-? 

(2) 

J 

0r4]           ?rr4 

=  4if8' 

(3) 

J 

0r4~|           Trr4 

(4) 

(b)   Since  for  the  circle  the  moments  of  inertia  about  all 

diameters  are  equal, 

Ix  +  Iy=2Ix  =  J  = 

—  ■      •      7=7= 

2    ,        ..        lx         ly 

7rr4 
4 

Trd4 
"  64* 

(5) 

For  a  circular  quadrant,  Ix  =  Iv  =  ~  J  = 

7rr4 
=  16* 

(6) 

For  a  semicircle, 

I     -I     ~X-J- 

7rr4 
=    8" 

(7) 

4.   Ellipse: 

Iy=       Cx2dA     = 

—  fax2(a2-x2)* 

a  J -a 

dx  = 

ira3b 
~4~' 

(1) 

Ix=Jy*dA  = 

^fb/(b2-y2)*dy  = 

irab3 
~4~' 

(2) 

J  =  Ix  +  Iy  = 

T-f(a2  +  V). 

(3) 

187.  Moment  of  Inertia  of  Compound  Areas,  —  Since 
the  moment  of  inertia  is  always  positive,  the  moment  of 
inertia  of  an  area  about  any  axis  is  equal  to  the  sum  of  the 


MOMENT  OF  INERTIA  OF  COMPOUND  AREAS     353 

moments  of  inertia,  about  that  axis,  of  the  parts  into  which 
the  area  may  be  divided.  In  some  cases  the  area  being 
considered  the  difference  of  two  areas,  its  moment  of  inertia 
will  be  equal  to  the  difference  of  the  moments  of  inertia  of 
the  two  areas. 

Example  1 .  —  Find  the  moment   _ 
of  inertia  of  the  T-section  shown 
with  the  dimensions  on  figure. 

(a)    About    the    axis    of    X 
through  top  of  section. 

(6)   About  the  axis  X0  through    - 
the  center  of  gravity. 

(a)  7I  =  p/i3  =  J(4-l)l3  +  Hlx43)  =  H-^=¥(ms.)4. 

XAy      3X|+4X2 


K -a 

i"*714 

* Xr 


(b)  2/o  = 


9i       19. 

7  =  l4ms- 


A  3  +  4 

7o  =  /*-Ai/o2  =  -y— 7(H)2  =  22.33-9.32=  13.01  (ins.)4. 

b 


Example    2. 
tion: 

hollow  square. 


Hollow    rectangular    sec- 
&!4)    for 


^(Vh-b^h). 


Example  3.  —  Hollow  circular  section 


I  —  t  (V24  —  ri4) ,  about  a  diameter ; 

7T 

J  =  9  (r24  —  ri4) ,   about  axis  through 
center. 
Or  in  terms  of  the  diameters: 


64 

7T 

32 


W  -  d,*). 


354 


INTEGRAL  CALCULUS 


Example  4.  —  Find  the  moment  of  inertia  of  the  section 
shown. 

l*  -  T2W  ~  2(?+f  °2  -  2aT  (fc))'  b^  »■  Art' 183 

=  ^  (6)4  -  ^  -  t  (2)2  (3)2  +  2.3-4^=  46.33  (ins.)4. 

h  =  I* 

j  =  Ix  +  ly  =  92.66  (ins.)4. 


Example  5.  —  Find  the  moment  of  inertia  of  the  trapezoid 
(a)  about  its  lower  base;   (b)  about  the  gravity  axis. 

t        Itu,    ltU      4  .  63  .  8  •  63 
J.-4W  +  I2W-— +  -12- 

-  216  +  144  =  360  (ins.)4. 
IG  =  IX-  Aa2  =  360  -  36  (§)2  =  104  (ins.)4. 


CHAPTER  VII. 

APPLICATIONS.     PRESSURE.     STRESS. 
ATTRACTION. 

188.  Intensity  of  a  Distributed  Force.  —  A  distributed 
force  is  one  that  acts  on  a  surface,  such  as  the  pressure  of 
water  against  the  surface  of  contact,  the  pressure  of  a  weight 
upon  the  surface  of  its  support;  or,  one  that  acts  through  a 
given  volume,  such  as  the  attraction  of  the  earth  on  a  body. 

All  forces  are  really  distributed  forces  since  no  finite  force 
can  act  at  a  point  of  no  area;  although  this  is  true,  in  some 
cases  it  is  convenient  to  regard  a  force,  whose  place  of  appli- 
cation is  small,  as  though  it  were  applied  at  a  point.  Such 
a  force  is  called  a  concentrated  force.  A  distributed  force  is 
conceived  as  " equivalent  to"  a  concentrated  force  called 
the  resultant  force,  when  the  force  of  gravity  acting  on  every 
particle  of  a  body  is  taken  as  acting  at  a  point  within  the 
body,  called  the  center  of  gravity.  A  distributed  force  is 
regarded  as  the  limiting  case  of  a  system  of  concentrated 
forces  whose  number  becomes  larger  as  their  individual 
magnitudes  become  smaller.  It  is  thus  that  a  force  is  re- 
garded as  having  a  definite  point  of  application  and  a  definite 
line  of  action:  when  so  regarded  it  is  a  localized  vector  quantity. 
When  a  force  is  distributed  over  an  area,  the  intensity  of  the 
force  at  a  point  is  the  number  of  units  of  force  acting  on  a 
unit  of  area  including  that  point. 

Briefly  the  intensity  is  defined  as  the  force  per  unit  of 
area.  If  the  force  is  uniformly  distributed,  the  intensity  p 
will  be  equal  to  the  force  P,  acting  on  the  entire  area,  divided 
by  the  area  A ;  that  is, 

355 


356  INTEGRAL  CALCULUS 

If  the  force  is  not  uniformly  distributed,  the  intensity  at 
any  point  of  the  area  will  be  given  by 

'-&Ih1-S 

the  limit  of  the  ratio  of  the  force,  acting  on  a  small  element 
of  the  area,  to  that  element  as  it  approaches  zero  as  a  limit. 
When  the  intensity  varies  from  point  to  point  over  any  area, 
the  force  on  that  area  divided  by  that  area  gives  the  average 
intensity  on  the  area.  In  any  case  the  entire  force  is  given 
by 

(3) 


=  j  pdA, 


where  p,  if  variable,  must  be  expressed  in  the  same  terms  as 
dA  in  order  to  get  P  by  integration. 
If  p  is  constant, 

P ;  =  p   I  dA  =  pA,    and    P  ~  ~\'  (1) 

189.  Pressure  of  Liquids.  —  The  pressure  of  a  liquid 
on  a  surface  is  normal  to  the  surface,  and  the  intensity  of 
pressure  varies  as  the  depth  of  the  point  below  the  free 
surface  of  the  liquid.     The  intensity  is  given  by 

p  =  wh,  (1) 

where  w  is  the  weight  of  a  cubic  unit  of  the  liquid  and  h, 
called  the  head,  is  the  depth  of  the  point  below  the  free 
surface. 

If  w  is  expressed  in  pounds  per  cubic  foot,  h  should  be  in 
feet,  and  p  will  then  result  in  pounds  per  square  foot.  For 
water  w  is  usually  taken  as  62^  lbs.  per  cubic  foot,  and  the 
intensity  of  pressure  given  in  pounds  per  square  foot.  When 
the  intensity  p  is  constant  on  any  horizontally  immersed 
plane  surface  the  total  pressure  P  is,  by  (1)  Art.  188, 

p  =  Awh  =  62.5  Ah  lbs.  (2) 


PRESSURE  OF  LIQUIDS  357 

When  the  surface  under  pressure  is  not  horizontal,  by  (3), 
Art.  188, 

wxdA,  (3) 


where  the  limits  of  x  are  the  least  and  greatest  heads  on  the 
area.  When  the  area  extends  to  the  surface  of  the  liquid, 
the  lower  limit  becomes  zero  and  the  upper  may  be  taken 
as  h. 

Since  in  (3),       fxdA  =  xA,     by  (2),  Art.  175, 
wxdA  =  Awx  =  62.5  Ax  lbs. 


/ 


Hence,  the  total  pressure  on  an  immersed  area  is  the  product  of 
that  area,  the  weight  of  a  cubic  unit  of  water,  and  the  head  upon 
its  c-enter  of  gravity. 

In  general,  the  pressure  of  any  liquid  upon  an  area  is  equal 
to  the  weight  of  a  column  of  liquid  whose  base  is  the  area  pressed 
and  whose  height  is  the  depth  of  the  center  of  gravity  of  the  area 
below  the  surface. 

Example  1.  —  The  vertical  face  of  a  dam  subjected  to  the 
pressure  of  water  is  h  ft.  in  height  and  b  ft.  in  breadth.  The 
pressure  of  the  water  varies  as  the  depth;  the  intensity  at 
a  depth  x  is  wx,  w  being  the  constant  weight  of  a  cubic  unit 
of  water.  Required  the  total  pressure  on  the  face  of  the 
dam,  and  the  location  of  the  center  of  pressure. 

Let  the  area  of  pressure  be  divided  into  strips  of  width  Ax 
and  length  b,  then  wx  •  b  Ax  is  approximately  the  pressure  on 
the  element  of  area  —  for  wx  is  the  intensity  of  pressure  at 
the  top  of  the  strip. 

The  sum  of  a  finite  number  of  terms  of  the  form  wbx  Ax 
would  give  a  result  for  the  total  pressure  less  than  the  actual 
value;   but  the  exact  value  is 

t,     ,.      ^A    7     a  i   Ch    j       wbx2~\h     wbh2     wh   ,.    ,  . 

P=  hm  V  wbxlx  =  wb  I    xdx=  -=-     =—=-  =  -=- -Oh.  (1) 

Ax  =  0^0  «/0  Z     JO  Z  Z 


358 


INTEGRAL  CALCULUS 


The  intensity  of  pressure  is  a  uniformly  varying  force  having 
zero  value  at  the  surface  of  the  water  and  value  wh  at  the 
bottom.  The  center  of  pressure,  being  the  point  of  applica- 
tion of  the  resultant  pressure,  is  given  by  taking  the  moment 


xS/3h 


of  P  about  the  surface  line  equal  to  the  limit  of  the  sum  of 
the  moments  of  the  elementary  pressures  about  that  line : 

wbx3~]h      wbh? 


x-P 


s: 


wbx2  dx  = 


Jo 


wbx2  dx 


j; 


wbxdx 


wbh*/S      2 
wbh2/2  ~  3 


(2) 


In  general,  the  center  of  pressure  of  a  rectangle  with  a 
side  at  the  surface  is  two-thirds  the  height  of  the  rectangle 
below  the  surface.  When  the  top  of  the  area  is  hi  below  the 
surface  and  the  bottom  is  h2  below,  the  total  pressure  is 


xdx  =  ~pr  (h22  —  hx2), 


and 


wb 


x  = 


hi 


x2dx 


wb 


J  ht 


dx 


I 

M~~ 

X6 

3_ 

=  x2~ 

ft,        2  W  -  ^!3 

*     3  W  -  h2 

2. 

hi 

(3) 


PRESSURE  OF  LIQUIDS 


359 


Note.  —  That  P 


It  may  be  noted  that  the  second  moment  in  the  numerator 
is  the  moment  of  inertia  of  the  area,  and  the  first  moment  in 
the  denominator  is  the  statical  moment. 

J  wbx  dx  in  (1)  is  the  reversal  of  a 

rate  may  be  seen  by  considering  the  rate  of  change  of  the 
total  pressure  when  the  depth  x  is  increased  by  Ax,  for  then 
the  pressure  P  on  the  area  is  increased  by  AP  =  wx  •  b  Ax, 
approximately,  and  AP/Ax  =  wbx  (nearly). 


r      AP      dP 

lim  -= —  =  -;—  =  wbx, 
ax=o  Ax       dx 


Hence 

the  rate  of  change  of  P;   and  P 


wbx  dx  as  in  (1);  so 


the  total  pressure  is  a  function  of  h  and  its  rate  of  change 
is  wbx  =  pb,  where  p  =  wx  is  the  pressure  on  a  unit  of 
area. 

Note.  —  Whenever  an  external  force  acts  on  a  body  it 
induces  a  resisting  force  within  the  body.  This  is  in  accord- 
ance with  Newton's  third 
law  of  motion.  This  in- 
ternal resistance  is  due  to 
the  molecular  forces  or 
stresses  within  the  body. 
A  stress  is  a  distributed 
force  acting  on  a  surface 

Example  2.  —  A  verti- 
cal rectangular  section 
ABBiAi  of  a  beam  of 
breadth  b  and  depth  d  is 
subjected  to  a  stress  of 
tension  and  compression  uniformly  varying  in  intensity  from 
zero  at  the  middle  fiber  to  St  and  *SC  at  the  outside  fibers,  at 
the  distance,  yx  =  \  d,  from  the  neutral  axis  or  middle  fiber 
of  the  section. 


360  INTEGRAL  CALCULUS 

Find  the  total  tensile  and  compressive  stresses  and  the 
centers  of  stress  on  each  half  of  the  section. 

The  intensity  of  stress  at  the  distance  y  from  the  neutral 
axis  is  S/yitjj  hence  S/yiyb  Ay  is  approximately  the  stress  on  a 
strip  Ay  in  depth  —  the  intensity  at  the  edge  of  the  strip 
being  taken.  The  sum  of  a  finite  number  of  such  terms 
would  give  a  result  less  than  the  total  stress  on  the  half 
section;  but  the  exact  value  is  given  by 

p  =  hm2,  -yb^y  =  —l  ydy 

d 

Sb  ?/2>     Sb     f      Sbd 

For  the  center  of  stress,  the  point  of  application  of  the  total 

stress, 

d 

-    t>       CmSh  2.7        Sby^     Sb    2T2      Sbd*. 

-_Sbd2  /Sbd_  1 
"      y~    12  /     4    ~3d' 

When  S  =  St  =  Sc,  P  —  Pi]  hence,  the  total  tensile  and 
compressive  stresses  form  a  couple  with  arm  §  d,  the  moment 
being 

—t-  •  ~d  =  — ^— ,  called  the  section  modulus. 

By  the  mechanics  of  beams,  if  M  denote  the  moment  of  the 

external  forces  acting  on  the  beam,  M  =  — ^ — 

Example  3.  —  A  vertical  circular  section  of  a  beam  is 
subjected  to  a  stress  of  tension  and  compression  uniformly 
varying  in  intensity  from  zero  at  the  horizontal  diameter 
2  a  of  the  section  to  St  and  $c  at  the  top  and  bottom  fibers, 
respectively. 


PRESSURE  OF  LIQUIDS 


361 


Find  the  total  stresses  on  the  upper  and  lower  semicircles 
and  the  centers  of  stress  on  the  semicircles. 


Denoting  by  P  the  total  stress  on  either  semicircle,  and 
taking  St  =  Sc  =  S; 

P=\imy\a-y2xAy  =  —  V  ^a2-y2ydy 

Ay=0  ^0  CL  d    Jo 

=  2S    Tra4  =  Swa*      Sird3 . 
a  "  16         8  64  ' 

-  =  Srd?    /Sd2  ^Zird 

"     y        64  /     6    ==   32  ' 

Hence   the   couple   formed   by  the  forces  P  has  an  arm, 
2y  =  Ts^d,  and 

M  =  ^Sd2  '—ird  =  -7^-,  the  section  modulus. 

O  It)  oZ 

Example  4.  —  Find  the  total  water  pressure  upon  the  end 
of  a  circular  right  cylinder  immersed  lengthwise,  one  element 
of  the  cylinder  just  at  the  surface  of  the  water.  Find  the 
center  of  pressure  of  the  circular  area. 


362 


INTEGRAL  CALCULUS 


The  intensity  of  pressure  at  a  depth  y  being  wy,  the 
approximate  pressure  on  a  strip  is  wy  •  2  x  A?/. 
o 


The  total  pressure  is  P  =  lim  ^    2  wz?/  A?/ 


Ay=0 


(2ay  -  y2Yydy  =  2wa  I      (2 ay  -  y2)*dy 

o  Jo 


=  2toa~  =  W7ra3;     (Ex.  13,  Exercise  XXV.) 

f*2a  r*2a 

2w  I      (2ay-y2)y2dy     $a-2w  I      (2ay-y2)*ydy 

—  Jo Jo 

y  "  P  ~  WTTO? 

=  ^4-  =  7«  =  |d.     (Ex.  13,  Exercise  XXV.) 
wira?       4         8 


EXERCISE   XLI. 

1.  (6)  The  pressure  upon  one  side  of  the  gate  of  a  dry  dock,  the 
wetted  area  being  a  rectangle  80  ft.  long  and  30  ft.  deep,  is  to  be  found 
exactly.     Take  w  =  62£  lb.  for  the  weight  of  a  cubic  foot  of  water. 

(c)   Find  the  depth  of  the  center  of  pressure.  Ans.  (c)  20  ft. 

(a)   Find  the  pressure  approximately  by  a  limited  number  of  terms. 


(See  Art.  154.) 


Ans.  (b)  112^  tons. 

2.  The  pressure  on  the  gate  that  closes 
a  water  main  half  full  of  water,  the  diam- 
eter of  the  main  being  8  ft.  Get  the 
exact  (6)  pressure  only,  (c)  Find  the 
center  of  pressure. 

Ans.  (b)  P  =  1^w  lbs.    (c)  y  -  |»  ft, 

3.  Find  the  exact  pressure  on  a  cir- 
cular disk  10  ft.  in  diameter,  submerged 
below  water  with  its  plane  vertical  and 
its  center  10  ft.  below  the  surface.    Here 


ATTRACTION.      LAW  OF  GRAVITATION  363 

p  =  w  C=a     (10  -y)  2xdy  =  2w  f°  (10  -  y)  (a2  -  yrf  dy 

J— 5=— a  •/— 5  =  — a 

=  20 16' f  (a2  -  y2)h dy-2wj   (a2  -  y2) * y cfy 

=  20 w;  •  -2/  7T  +  |  (a2  -  ?/2)fl  =  250ttw 

J— 5  =  — a 

=  2507r-62ilb. 

4.  Find  the  pressure  on  the  face  of  a  temporary  bulkhead  4  ft.  in 
diameter  closing  an  unfinished  water  main,  when  water  is  let  in  from 
the  reservoir.  The  center  of  bulkhead  is  40  ft.  below  the  surface  of  the 
water  in  the  reservoir.  Ans.  Nearly  16  tons. 

6.  Find  the  pressure  on  the  end  of  a  parabolic  trough  when  it  is  full 
of  water.  The  parabola  has  its  vertex  downward,  the  latus  rectum  is 
in  the  surface  and  is  4  ft.  long.     Here 

P  =  lim  V   2  w  (1  -  y)  x  Ay  «=  2  w  C  2  (1  -  y)  y*  dy 
=  4  w  (    y*dy  -  f  y*  dy 
=  4  w  [f  j/*  -  \  2,1]*  =Hw  =  66|  lbs. 


jr 

Y 
--4- 

" "1       x 

Ij, 

jy^« 

sV^s^^//j£^ 

^ ijf 

6.  A  horizontal  cylindrical  tank  is  half  full  of  oil  weighing  50  lb.  per 
cubic  foot.  The  diameter  of  each  end  is  4  ft.  Find  the  pressure  on 
each  end.     Find  the  pressure  when  the  tank  is  full  also. 

Ans.  266f  lb.;  1256  1b. 

190.  *  Attraction.  Law  of  Gravitation.  —  Every  portion 
of  matter  acts  on  every  other  portion  of  matter  with  forces 
of  attraction  or  repulsion.  According  to  Newton's  Law  of 
Universal  Gravitation,  every  particle  of  matter  attracts 
every  other  such  particle  with  a  force  which  acts  along  the 
line  joining  the  two  particles,  and  whose  magnitude  is  pro- 

*  This  article  is  based  on  a  discussion  in  Fuller  and  Johnston's 
Applied  Mechanics. 


364  INTEGRAL  CALCULUS 

portional  directly  to  the  product  of  their  masses  and  in- 
versely to  the  square  of  the  distance. between  them. 

If  the  masses  of  the  particles  are  m  and  m,\  and  the  distance 
between  them  is  r,  the  law  may  be  expressed  algebraically  by 

F-K**,  (1) 

where  F  is  the  attractive  force  between  the  particles  and 
K  is  a  constant,  determined  by  experiment,  its  numerical 
value  depending  on  the  units  in  which  F,  m,  mi  and  r  are 
expressed.  The  value  of  K  having  been  determined  in  one 
case  is  then  known  for  all  cases. 

While  formula  (1)  expresses  the  law  of  gravitation,  the 
general  algebraic  expression  for  the  law  of  attraction  would 
be 

F  =  Kwr  (2) 

where  <f>(r)  is  some  function  of  the  distance  between  the 
particles,  depending  on  the  nature  of  the  attractive  force, 
K  is  a  constant,  and  m  and  mi  other  quantities  than  the 
masses  of  particles. 

In  interpreting  formula  (1),  it  is  to  be  noted  that  it  applies 
strictly  only  to  particles;  for  the  particles  having  finite 
masses  must  have  finite  dimensions  and  hence,  as  the  distance 
between  them  is  diminished,  r  cannot  be  less  than  a  certain 
finite  quantity  and  the  maximum  value  of  F,  when  the 
particles  are  in  contact,  will  be  a  finite  quantity.  If  r  were 
taken  to  be  zero  in  any  case,  F  for  finite  values  of  m  and  m\ 
would  become  oo  which  would  be  impossible  under  the 
conditions. 

The  formula,  while  applying  strictly  only  to  particles, 
gives,  to  a  close  approximation,  the  attraction  between  two 
bodies  of  finite  size,  whose  linear  dimensions  are  small 
compared  to  the  distance  between  them.  In  the  application 
of  the  law  the  attraction  of  one  particle  on  another  may  be 


ATTRACTION.      LAW  OF  GRAVITATION  365 

regarded  as  acting  at  a  point.  It  will  be  shown  that  any 
sphere  attracts  any  outside  particle  as  if  the  whole  attraction 
was  towards  a  point  at  the  center  of  the  sphere,  but,  in 
general,  the  attraction  of  bodies  on  exterior  particles  is  not 
always  towards  the  center  of  gravity  of  the  attracting  body. 

Attraction  of  gravitation  is  a  mutual  action  between  two 
particles  or  bodies;  that  is,  each  exerts  an  attractive  force 
upon  the  other,  the  two  forces  being  equal  in  magnitude  and 
opposite  in  direction.  This  is  implied  in  the  Law,  and  it  is 
also  in  accordance  with  the  law  of  "action  and  reaction," 
Newton's  third  law  of  motion. 

It  is  evident  that,  in  formula  (1),  K  is  equal  to  the  force 
with  which  two  particles  of  unit  mass  at  a  unit  distance 
apart  attract  each  other. 

If  the  equation  is  divided  by  mi,  then 

—  =  a  =  K-^j  (3) 

mi  r2 

where  a  is  the  acceleration  which  would  be  produced  in  the 
mass  mi  by  the  attraction  of  the  mass  m  at  a  distance  r. 

The  quantity  K  —2  would  also  equal  the  force  of  attraction 

exerted  by  the  mass  m  on  a  mass  unity  at  a  distance  r. 
Briefly  this  is  called  the  attraction  at  the  point,  at  which  the 
unit  mass  is  situated,  exerted  by  the  mass  m. 

The  attraction  at  a  point  exerted  by  any  mass  is  called  the 
strength  of  the  field  of  force,  or  briefly,  the  strength  of  field,  by 
which  the  space  through  which  the  attraction  of  the  mass  is 
exerted  is  expressed. 

Electrostatic  and  magnetic  attraction  and  repulsion  are 
other  examples  of  forces,  which  are  governed  by  laws  similar 
to  that  of  gravitation. 

The  following  examples  are  based  on  the  law 

F-K±, 

r 


366 


INTEGRAL  CALCULUS 


where  F  is  the  attraction  of  a  particle  of  mass  m  for  a  par- 
ticle of  unit  mass,  the  body  being  taken  as  homogeneous,  of 

uniform  density;  that  is,  each 
cubic  unit  having  the  same 
weight. 

Example  1.  —  Attraction  of  a 
Rod  of  Uniform  Section.  —  (a) 
Let  the  rod  of  small  section  be 
in  the  form  of  a  circular  arc; 
to  find  the  attraction  at  the 
center  of  the  circle. 

Let  r  be  the  radius,  a  the 
angle  subtended  at  the  center, 
and  m  the  mass  of  a  unit  length 
of  the  rod.  Take  the  axis  OX 
bisecting  the  angle  a,  and  let  0 
be  the  angle  which  the  radius 
to  any  point  P  makes  with  OX.  The  attraction  at  0  of  a 
particle  at  P  is 


AF 


Km  As      Km  Ad 


(1) 


r*  r 

Since  all  the  elementary  forces  of  attraction  are  directed 
to  the  point  0,  the  resultant  R  is  found  from  the  sum  of  the 
components  of  the  elementary  forces. 


Xx  =  Kpf 


cos  Bad  = sin-> 

«       *  r  2 

2 


a 

2 

the  attractions  being  neutralized. 
Hence, 

When  a  =  ir,    R  = 


2  Km 


2Km  .    a 

sin-- 

r  2 


(2) 

(3) 


ATTRACTION.      LAW  OF  GRAVITATION  367 

When  a  =  2t,  R  =  0;  since  the  arc  being  a  circumfer- 
ence of  a  circle,  the  attractions  neutralize  each  other. 

(6)  Let  the  rod  be  straight;  to  find  the  attraction  at  a 
point.  Let  r  be  the  shortest  distance  from  a  point  0  to  the 
rod.  Taking  0  as  origin,  the  equation  of  the  rod  is  x  =  r 
(constant) . 

When  the  rod  is  B'AB  the  angles  may  be  taken  as  in  (a) 
for  the  circular  arc.  The  attraction  at  0  of  a  particle  at  P' 
on  the  rod  is 

.  ,-,        Km    4         Km  cos2  6A  /1A 

AF  =  wy  y  =  — ? —  y'  (  } 

The  resultant  attraction  is  found  as  in  (a) ;  since  y  =  r  tan  6, 

rdd 

Xx  =  ^f  comedy  =  ^f  cosede 


2  Km  .    a 
2 


sin  - ,  as  in  (a) . 


£  Y  =  ^JcosHsmOdy  =  ^p  f*  sin  add  =  0. 

_  2 

Hence,        #  = sin-,  as  in  (a).  (2') 

r  J 

It  is  thus  shown  that  if  the  straight  rod  B'AB  is  of  the 
same  mass  per  unit  length  as  P2Pi,  the  resultant  attraction 
of  B'AB  at  0  is  the -same  as  the  attraction  of  P%Pi,  since  the 
sum  of  the  attractions  of  the  elementary  masses  m  Ay  and 
the  sum  of  the  attractions  of  the  elementary  masses  m  As 
have  the  same  limit. 

(&')  Let  the  rod  be  still  straight  but  AXB,  the  angles  with 
OX  of  the  lines  from  0  to  the  ends  being  «i  and  a2;  then, 

V  X  =  —  pcos  6de  =  ~  (sin  a2  -  sin  a,), 

STr      Km   fa*  .    a  ln      Km  .  v 

Y  = I     sin  Odd  = (cos  ai  -  cos  a2). 
r    Jai                     r 


368  INTEGRAL  CALCULUS 

Hence, 


R 

r 

—  COS (aj 

~«l)J  = 

and 

tan0r 

%Y_ 

cos  a<i  — 
sin  a2  — 

COStti 

-%x~ 

sinai 

'      fi~ 

«2  +  Cki 

^sin^,     (30 


tan^-V 


2 

the  line  of  action  of  R  bisecting  the  angle  AxOB,  subtended 
by  AYB  at  0. 

(b")   Let  the  point  0  be  at  A,  making  r  =  0  and  (3') 
indeterminate.     Then, 

Kmdy 


AF  = 


r 


...  ^=Xmr^=wi_i.)=^fa-^)=gM?  (4) 

•/»,  y  VJi     2/2/  2/2?/i  2/22/1 

where  M  =  the  entire  mass  of  the  rod. 

If  the  point  0  is  taken  at  the  end  of  the  rod,  yi  =  0  and 
equation  (4)  gives  R  =  oo .  This  is  impossible;  for,  as 
stated  in  Art.  190,  r  cannot  be  zero  for  finite  particles.  If, 
however,  2/2  =  oo , 

R  =  ^, 

2/i 

making  R  a  finite  quantity  for  any  length  from  A\. 

If  the  point  0  were  taken  on  the  rod  between  A 1  and  B, 
with  lower  limit,  —yh 


R  =  -Km 


(L+1-) 

\2/i      2/2/ 


and,  if  0  were  taken  at  the  middle  point  of  the  rod,  it  is 
evident  that  R  =  0. 

Example  2.  —  Attraction  of  a  Spherical  Shell  at  a  Point.  — 
Find  the  resultant  attraction  of  a  spherical  shell  of  uniform 
density  and  small  uniform  thickness  on  a  particle  of  unit 
mass,  M'  being  the  mass  of  the  shell. 


ATTRACTION.      LAW  OF  GRAVITATION 


369 


(a)  Let  the  point  P  outside  the  shell  be  the  position  of 
the  particle. 

Let  7  be  the  density  and  t  the  thickness  of  the  shell,  0  its 
center,  and  a  the  radius  ON;  let  NP  =  r  and  OP  =  d.  If 
the  circle  be  revolved  about  OP  as  an  axis  through  an  angle 
2  7r,  a  thin  spherical  shell  of  thickness  t  will  be  generated,  and 
an  elementary  volume  will  be  generated  by  the  elementary 
area  at  iv,  whose  mass  will  be  AM'  =  yt  •  2  ira2  sin  0  Ad, 
approximately. 


The  attraction  of  the  elementary  mass  at  N  for  the  particle 
of  unit  mass  at  P  is 

Kyta  Ad 


A2F 


(1) 


This  attraction  may  be  resolved  at  P  into  a  component  X 
along  PO  and  a  component  Y  perpendicular  to  PO.  To 
every  elementary  mass  at  iV  there  is  a  corresponding  mass 
at  N',  whose  attraction  at  P  is  X  along  PO,  and  —  Y  per- 
pendicular, which  neutralizes  Y.  Hence  the  attraction  of 
AM'  is  along  PO,  and  is  given  approximately  by 


XT1      Kyt- 2  ira2  sind  Ad 

AF  =  — ^ cos  <$>. 


(2) 


From  the  geometry  of  the  figure, 

r2  =  a2  +  d2-  2adcosd, 


370  INTEGRAL  CALCULUS 

which  differentiated  gives 

t  dv 
rdr  =  ad  sin  Odd:     .'.      sin  6 


and  from  the  figure,  cos  0  = 


ad-dd' 
d  —  a  cos  6 


r 
Substituting  these  values  in  (2)  gives  exactly, 


at  Id2  -  a2  +  r2> 
d2\         r2 


dF  =  Kyirf2{--^-)dr;  (3) 

hence,  F  =  Kyir  j2  J ^  ( ? )dr 


AKyiraH      KM' 
d2  d2 


(4) 


It  follows  from  (4)  that  the  attraction  is  the  same  as 
though  the  mass  of  the  shell  were  concentrated  at  its  center. 
It  follows  also  that  a  sphere,  which  is  either  homogeneous  or 
consists  of  concentric  shells  of  uniform  density,  attracts  a 
particle  without  the  sphere  as  if  the  mass  of  the  sphere  were 
concentrated  at  its  center.  This  law  holds  almost  exactly, 
for  bodies  slightly  flattened  at  the  poles,  if  the  particle  is  not 
too  close  to  the  attracting  body.  Since  both  these  con- 
ditions exist  in  the  case  of  the  Earth  and  other  members  of 
the  Solar  System,  this  law  has  important  applications. 

(6)  Let  the  point  P'  inside  the  shell  be  the  position  of  the 
particle.  The  equation  (3)  in  case  (a)  is  true  for  this  case 
too,  but  the  limits  for  r  are  now  a  —  d  and  a  -f-  d.     Hence, 

„      Kywat  [a2  —  d2 


\a2-d2  ,     >+"     n  ... 


that  is,  the  resultant  of  all  the  attractions  of  the  elementary 
masses  of  the  spherical  shell  on  a  particle  within  the  shell 
is  zero. 

(c)    Let  the  point  where  the  particle  is  be  on  the  surface 
of  the  shell. 


ATTRACTION.      LAW  OF  GRAVITATION  371 

In  (4)  making  d  =  a  gives 

F  =  4Kywt  =  ^—  (6) 

Corollary.  —  If  a  particle  be  inside  a  homogeneous  sphere 
at  a  distance  d  from  its  center,  all  that  portion  of  the  sphere 
at  a  greater  distance  from  the  center  than  the  particle  has 
no  effect  on  the  particle,  while  the  remaining  portion  attracts 
the  particle  in  the  same  way  as  if  the  mass  of  the  remain- 
ing portion  were  concentrated  at  the  center  of  the  sphere. 
Thus  the  attraction  of  the  sphere  on  the  particle  is 

„       %Kiryd*       IKiryd 


d- 


(7) 


that  is,  within  a  homogeneous  sphere  the  attraction  varies 
as  the  distance  from  the  center.  The  attraction  of  a  sphere 
of  mass  M  on  a  particle  at  the  surface  is  from  (7),  making 
d  =  a, 

F  = -sKirya  = -^--  (8) 

Hence,  the  attraction  for  an  external  particle  is 

'-¥■ 

where  d  is  the  distance  from  the  particle  to  center  of  sphere. 

Note.  —  The  propositions  respecting  the  attraction  of  a 
uniform  spherical  shell  on  an  external  or  internal  particle 
were  given  by  Newton  (Principia,  Lib.  I,  Prop.  70,  71). 

It  was  in  1685,  nineteen  years  after  he  had  conceived"  the 
theory  of  universal  gravitation,  that  he  completed  the  veri- 
fication of  the  theory,  by  proving  that  a  sphere  in  which  the 
density  depends  only  upon  the  distance  from  the  center 
attracts  an  external  particle  as  if  the  mass  of  the  sphere  were 
concentrated  at  its  center.  Thus  was  the  great  induction 
by  this  supplementary  proposition  finally  established. 


372  INTEGRAL  CALCULUS 

Example  3. — Attraction  of  the  Earth.*  —  I.  Find  the 
relation  between  the  attraction  of  the  Earth  on  a  body  at 
the  surface  and  at  a  point  h  feet  above  the  surface. 

Taking  the  Earth  as  a  sphere  whose  density  is  a  function 
of  the  distance  from  the  center,  R  as  the  radius,  and  F  and  F' 
as  the  Earth's  attraction  upon  the  body  at  the  surface  and 
at  h  feet  above  the  surface, 

F/F'  =  (R  +  hy/R2     (by  Ex.  2,  (8)  and  (9)), 
or  F'  =  FR2/(R  +  h)2.  (1) 

If  h  is  a  small  fraction  of  R,  then  approximately, 

F'  =  F  (1  +  h/R)~2  =  F  (1  -  2  h/R).  (2). 

Since  the  "weight"  of  a  body  is  the  force  with  which  the 
Earth  attracts  it,  the  equations  (1)  and  (2)  give  the  relation 
between  the  weight  of  a  body  at  the  surface  and  at  a  height 
h  feet  above  the  surface.  And,  if  g  and  g'  are  the  values  of 
the  acceleration  of  gravity  at  the  surface  and  at  the  point  h 
feet  above  the  surface,  since  F/F'  =  g/g',  the  equations  give 
the  relation  between  g  and  g'  also. 

(a)  Find  approximately  at  what  height  above  the  surface 
will  the  weight  of  a  body  be  TV  of  one  per  cent  less  than  at 
the  surface. 

Taking  the  mean  radius  of  the  Earth  as  20,902,000  ft., 

F'/F  =  l-2h/R  =  1-  1/1000; 

.         R        20,902,000       1AilRf",    , 
"     *  =  2005=       2000       =10>451feet- 

Corollary.  —  A  mass  which  at  the  surface  weighs  one  pound 
at  10,451  ft.  will  weigh  0.999  lb. 

(b)  Find  how  much  the  value  of  g  is  changed  by  a  change 
of  elevation  of  one  foot  above  the  surface. 

*  This  example  is  based  on  examples  in  Hoskins's  Theoretical  Me- 
chanics. 


ATTRACTION.      LAW  OF  GRAVITATION  373 

The  value  of  g  for  different  latitudes  and  elevations  is  given 
by  the  following  formula,  in  which  g  is  in  feet  per  second,  I 
is  the  latitude,  and  h  the  elevation  in  feet  above  sea  level : 

g  =  32.0894  (1  +  0.005243  sin2 1)  (1  -  0.000.0000957  h). 
This  gives 

g  =  32.0894  at  the  equator  at  sea  level,  and 

g  =  32.174  at  45°  latitude  at  sea  level; 

this  latter  value,  g  =  32.174  ft.  per  sec.  per  sec.  is  the 
standard  value. 

II.  Find  the  relation  between  the  attraction  of  the  Earth 
on  a  body  at  the  surface  and  at  a  point  h  below  the  surface, 
(a)  Taking  the  Earth  as  a  sphere  of  uniform  density  of  radius 
R, 

F"/F  =  (R-  h)/R  =  1  -  h/R     (by  Cor.  Ex.  2),      (3) 

where  F  and  F"  denote  the  attraction  at  the  surface  and  at 
h  below  the  surface. 

Corollary.  —  Under  these  conditions,  the  weight  of  a  body 
and  the  value  of  g  would  decrease  with  the  depth  h  below  the 
surface. 

(b)  Taking  the  Earth  as  a  sphere  whose  density  is  a 
function  of  the  distance  from  the  center,  let  y  denote  the 
mean  density  of  the  whole  Earth  and  70  the  mean  density 
of  the  outer  shell  of  thickness  h. 

Let  M  be  the  mass  of  the  whole  Earth,  M"  that  of  the 
inner  sphere  of  radius  R  —  h,  m  the  mass  of  the  attracted 
body,  F  and  F"  the  attraction  at  the  surface  and  at  h  below 
the  surface.  Then  F  is  equal  to  the  attraction  between  two 
particles  of  masses  M  and  m  whose  distance  apart  is  ~R,  and 
F"  is  equal  to  the  attraction  between  two  particles  of  masses 
M"  and  m  whose  distance  apart  is  R  —  h.     That  is, 

F  =  KMm/R2,    F"  =  KM"m/(R  -  h)2; 
.  F"      M"(    R    \« 

hence>  T  =  ir{R=h)'  (4J 


374  INTEGRAL  CALCULUS 

Now    M  =  t7r#37;    M  -M"  =  J 7r7o  [Rs  -  (R  -  h)*]; 

•■•  ¥ =>-?H^)H-?)+?(t)'  <» 

which  substituted  in  equation  (1),  gives 

£=('-?)G^H-(V>     « 

If  /i  is  a  small  fraction  of  #,  equation  (6)  may  be  reduced 
to  the  approximate  formula, 

Corollary.  —  If  the  mean  density  of  the  outer  layer  of  the 
Earth  is  less  than  two-thirds  the  mean  density  of  the  whole 
Earth,  the  weight  of  a  body  increases  as  it  is  taken  below 
the  surface  of  the  Earth.     (See  Ex.  1,  Art.  171.) 

The  mean  density  of  the  Earth  being  taken  as  5.52  and 
that  of  the  layer  near  the  surface  as  2.76,  about  the  density 
of  the  rocks,  makes  70/7  =  \  and  equation  (7), 

F"/F  =  W"/W  =  q"/q  =  1  +  h/2R.  (8) 

That  the  weight  of  a  body  increases  as  it  is  taken  below  the 
surface  has  been  shown  by  actual  trial.  From  (8),  the  depth 
to  which  a  body  must  be  taken  in  order  that  it  gain  T<^  of 
one  per  cent  in  weight  is  approximately, 

,       2X20,902,000       .1Qnn 

h=     10,000     =4180ft-; 

that  is,  a  mass  weighing  a  pound  at  the  surface  will  weigh 
1.0001  lb.  at  a  depth  of  4180  ft,  below  the  surface. 

Compared  with  case  I,  it  may  be  seen  that,  under  the 
conditions,  for  the  same  value  of  h,  the  gain  in  weight  is  one- 
fourth  as  much  as  the  loss  in  weight  when  the  body  is  above 
the  surface,  the  same  ratio  of  change  applying  to  the  value 
of  g  also. 


VALUE  OF  THE  CONSTANT  OF  GRAVITATION      375 

191.  Value  of  the  Constant  of  Gravitation.*  —  From  the 
foregoing  as  to  the  attraction  of  a  sphere,  it  follows  that  the 
formula  for  the  attraction  of  two  particles, 

F  =  Krm^_        [(1)Art<19o] 

will  apply  to  two  spheres,  which  are  either  homogeneous 
throughout  or  composed  of  a  series  of  concentric  shells,  each 
one  of  which  is  of  uniform  density,  m  and  m'  being  the  masses 
of  the  spheres  and  r  the  distance  between  their  centers. 

By  measuring  the  force  of  attraction  between  two  spheres 
of  known  mass  and  distance  apart,  the  value  of  K  the  constant 
of  gravitation  has  been  found.  As  stated  in  Art.  190,  its 
numerical  value  will  depend  on  the  units  used  for  the  other 
quantities  in  the  equation.  The  relation  between  the 
constant  K  and  the  mass  of  the  Earth,  taking  the  Earth  as  a 
sphere  whose  density  is  a  function  of  the  distance  from  the 
center  may  be  shown  as  follows. 

Let  the  units  be  the  British  gravitation  units,  and  let  R 
be  the  radius  of  the  Earth  in  feet,  M  its  mass,  y  its  mean 
density.  Consider  the  attraction  of  the  Earth  on  a  body  of 
mass  m  at  the  surface.  By  the  formula  (1)  of  Art.  190,  the 
value  of  the  attraction  is  KM?n/R2;  but  (since  the  unit  force 
is  the  weight  of  a  pound  mass)  expressed  in  pounds  force, 
its  measure  is  m.     Hence,  m  =  KMm/R2  or 

KM  =  R\  (1) 

Since  the  value  of  R  is  known,  either  K  or  M  can  be  found 
when  the  other  is  known.  Putting  for  M  its  value  in  terms 
of  7, 

K-UR*y  =  R2    or    ^7  =  ~-  (2) 

*^Arts.  191  and  192  are  based  on  Articles  in  Hoskins's  Theoretical 
Mechanics. 


376  INTEGRAL  CALCULUS 

Taking  y  =  345  lbs.  per  cu.  ft.  and  R  =  20,900,000  ft.  gives 
K  =  -r\-  =  3/(4 7T  X  20,900,000  X  345)=  3.31  X  10"11. 

4  7T/L7 

Otherwise,  if  the  value  of  K,  found  by  direct  measurement 
of  the  attraction  of  two  spherical  bodies,  is  substituted  in  (2) 
the  value  of  7,  the  mean  density  of  the  Earth  is  found  to 
be  5.527.  The  density  of  water  being  unity,  and  its  weight 
62.4  lbs.  per  cu.  ft.,  the  mean  density  of  the  Earth  is  about 
345  lbs.  per  cu.  ft.,  as  used  above. 

192.  Value  of  the  Gravitation  Unit  of  Mass.*  —  As 
stated  in  Art.  190,  the  force  with  which  two  particles  of  unit 
mass  at  a  unit  distance  apart  attract  each  other  is  equal 
to  K,  the  constant  of  gravitation;  this  is  evident  from  the 
equation, 

F  =  Rm^_       [(1)Art<  m] 

Let  m  pounds  be  the  mass  of  each  of  two  particles  which, 
when  one  foot  apart,  attract  each  other  with  one  pound  force. 
Substituting  K  =  3.31  X  10-11,  as  given  above,  putting 
F  =.  1,  m  =  mf,  and  r  =  1;  gives 

m  =  l/VE  =  173,800  lb. 

If  a  mass  equal  to  173,800  pounds  be  taken  as  the  unit 
mass,  the  constant  K  becomes  unity  and  the  formula  for 
attraction  is  then 

F  =  mm' 

The  gravitation  unit  of  mass  is  thus  shown  to  be  a  mass  equal 
to  about  173,800  pounds,  distance  being  in  feet  and  force  in 
pounds-force. 

193.  Vertical  Motion  under  the  Attraction  of  the  Earth. 
—  Let  the  Earth  be  taken  as  in  Example  3,  Art.  190,  r  as 
the  radius  and  s  the  distance  of  the  moving  particle  from 

*  See  Footnote  on  page  375, 


VERTICAL   MOTION 


377 


the  center  0.     Taking  distance,  velocity,  and  acceleration 
as  positive  outward,  then,  as  in  (1)  Ex.  3, 


F' 

F 

g 

Since 

dv 

di  = 

a  an 

gives 

v  dv  = 

ads. 

or     a'  = 


gr- 


-r  =  v,  eliminating  dt 


Hence,   a  =>  —  g'  =  —  ^-,  neglecting  air  re- 

S" 

sistance,  which  gives 

I    vdv=   j    —  gr2s~2ds; 

v2  lr  /l       1\ 

integrating,  ^  =  ^s~l  I  =  ^  [j  ~  J )  J 

then,  v2  =  2gr2^--^j 


(1) 


gives  the  velocity  of  a  particle  towards  the  Earth  from  any 
distance  s. 

For  the  velocity  acquired  by  a  body  in  falling  to  the  surface 
from  a  height  h,  put  s  =  r  +  h  in  (1),  giving, 


=  2K^h) 

=  2gh,  approximately, 


(2) 


if  h  is  small,  which  is  the  formula  when  g  is  constant,  as  it  is 
taken  near  the  surface.     When  -  is  small,  putting 


(h/r)'  +•••].       (3) 


r  + 
(2)  becomes 

v2  =  2  gh  [1  -  (h/r)  +  (h/r)2 


378  INTEGRAL  CALCULUS 

By  taking  any  number  of  terms  of  this  series,  an  approxi- 
mate result  may  be  gotten  as  nearly  correct  as  desired.  If, 
in  (1),  s  =  oo ,  v  =  V2  gr;  so  if  a  body  fell  toward  the  Earth 
from  an  infinite  distance,  its  velocity,  neglecting  air  resist- 
ance, would  be  V2  gr  =  6.95  miles  per  second,  for  r  =  3960 
miles.  If  falling  from  a  finite  distance  s,  the  velocity  must 
be  less  than  this.  Hence,  a  body  can  never  reach  the  earth 
with  this  velocity;  and  if  air  resistance  is  considered,  the 
velocity  for  s  =  co  is  less  than  V2  gr.  If  projected  outward 
with  velocity  V2  gr  and  air  resistance  be  neglected,  the  body 
would  go  an  infinite  distance.  This  velocity  is  called  the 
critical  velocity  or  velocity  of  escape,  for  under  the  conditions 
it  is  supposed  that  certain  particles  of  the  atmosphere  may 
escape  from  the  attraction  of  the  Earth. 

In  this  connection,  it  is  to  be  recalled  that  due  to  the 
Earth's  rotation,  there  is  at  its  surface  a  'centrifugal  force 
mg/289,  exerted  by  a  particle  of  mass  m,  which  lessens  the 
value  that  g  would  otherwise  have. 

194.*  Necessary  Limit  to  the  Height  of  the  Atmosphere. 
—  The  centrifugal  force  of  a  particle  of  mass  m  on  the  surface 

7710 

of  the  Earth  is  muftr  =  ^r,  and  at  a  distance  s  from  the 

7TIQS 

center  it  would  be  mo)2s  =      *„    .     The  Earth's  attraction  at 

289  r 

Trior2 
that  distance  being  — ^-,  in  order  that  the  particle  be  re- 

tained  in  its  path  these  two  forces  must  equal  each  other; 
mgs  _  mgr2 
'"'     289r  ~  -^~, 
or  s3  =  289  r3, 

hence  s  =  ^289  r  =  6.6  r 

=  26,000  miles  approximately; 
that  is,  a  height  above  the  surface  of  about  22,000  miles.     The 
actual  height  of  the  atmosphere  is  probably  much  less  than 

*  Bowser's  Hydromechanics. 


MOTION  IN  RESISTING  MEDIUM  379 

this.  The  estimates  of  the  height  by  various  scientists  have 
been  very  divergent  —  from  40  miles  to  216  miles;  but  the 
latter  appears  to  be  the  most  likely,  for  meteors  have  been 
observed  at  an  altitude  of  more  than  200  miles  and,  as  they 
become  luminous  only  when  they  are  heated  by  contact  with 
the  air,  this  is  evidence  that  some  atmosphere  exists  at  that 
height.  It  is  supposed  that  at  a  height  much  less  than  5.6  r, 
the  air  may  be  liquefied  by  extreme  cold. 

195.*  Motion  in  Resisting  Medium.  —  Consider  the 
motion  of  a  body  near  the  surface  of  the  Earth  under  the 
action  of  gravity  taken  as  a  constant  force  and  the  air  taken 
as  a  resisting  medium  of  uniform  density,  the  resistance 
varying  as  the  square  of  the  velocity. 

Let  a  particle  be  supposed  to  descend  towards  the  Earth 
from  rest,  and  let  s  be  the  distance  of  the  particle  from  the 
starting  point  at  any  time  t,  gk2  the  resistance  of  the  air  on 
a  particle  for  a  unit  of  velocity  —  gk2  being  the  coefficient 
of  resistance.     The  resistance  of  the  air  at  the  distance  s  from 

the  origin  will  be  gk2[-n)  ,  acting  upwards,  while  g  acts 

downwards,  the  mass  being  a  unit. 
The  equation  of  motion  is 

d2s  T  „  fds\2 

dt2 


9 


or  g  dt  = 

Integrating, 


<(f) 


I 


,        1,  dt 

gt  =  2kl0g—Js' 
1       hdt 

%  =  0,  v  =  0,  giving  C  =  0. 

*  Bowser's  Analytic  Mechanics.' 


380  INTEGRAL  CALCULUS 

Passing  to  exponentials, 

ds  _  1  ekot  —  e~kgt 
dt~  k  ekgt  +e~kgti 


(2) 


which  gives  the  velocity  in  terms  of  the  time.     To  get  it  in 
terms  of  the  space,  from  (1), 

.-.     log[l-fc2^y]=  -2gk\s  =  0,  v  =  0;d  =  0,      (3) 

f$-h* -**">•  (4) 


which  gives  the  velocity  in  terms  of  the  distance.  Also, 
integrating  (2); 

gkh  =  log  (ekgt  +  e~kgt)  -  log  2; 
/.     2  eQkH  =  efc(7<  +  e-*ffS  (5) 

which  gives  the  relation  between  the  distance  and  the  time 
of  falling  through  it. 

As  the  time  increases  the  term  e~kgt  diminishes  and  from 
(5)  the  space  increases,  becoming  infinite  when  the  time  is 
infinite;  but  from  (2)  as  the  time  increases  the  velocity 
becomes  more  nearly  uniform,  and  when  t  =  oo ,  the  velocity 
=  1/k;  and  although  this  state  is  never  reached,  yet  it  is 
that  to  which  the  motion  approaches. 

196.  Motion  of  a  Projectile.  —  If  a  body  be  projected 
with  a  given  velocity  in  a  direction  not  vertical  and  be  acted 
on  by  gravity  only,  neglecting  the  resistance  of  the  air,  it  is 
called  a  projectile.  The  path,  called  the  trajectory,  will  result 
from  a  combination  of  the  motions  due  to  the  velocity  of 
projection  and  to  g,  the  vertical  acceleration  of  gravity. 
Let  the  plane  in  which  a  particle  is  projected  with  a  velocity 


MOTION  OF  A  PROJECTILE 


381 


v  be  the  plane  of  XY,  and  let  the  line  of  projection  be  inclined 
at  an  angle  a  to  the  z-axis,  making  v  cos  a  and  v  sin  a  the 
resolved  parts  of  the  velocity  of  projection  along  the  axes  of 
x  and  y. 

Y 


Let  (x,  y)  be  the  position  of  the  particle  P  at  the  time  t; 
then,  since  the  horizontal  acceleration  is  zero  and  the  vertical 
acceleration  negative, 

d2x  _  d2y  _ 

dt?~    ;     ~df2~~g' 
Taking  the  first  and  second  integrals  of  these  equations, 
determining  the  value  of  the  constants  of  integration  corre- 
sponding to  t  =  0  and  t  =  t,  gives 


dx 


=  z;cosa; 


dy 


=  v  sin  a  —  gt; 


dt"w"~>      dt~w~"~      yv>  (1) 

x  =  vcos  at;       y  =  v  sin  at  —  |  gt2.  (2) 

Equations  (1)  and  (2)  give  the  coordinates  of  the  particle 
and  its  velocity  parallel  to  either  axis  at  any  time  t. 
Eliminating  t  between  equations  (2),  gives 


y  =  x  tan  a 


gr 


(3) 


2  v2  cos2  a' 

which  is  the  equation  of  the  trajectory,  and  shows  that  the 
path  of  the  particle  is  a  parabola. 
Putting  equation  (3)  in  the  form 


2  v2  sin  a  cos  a 
9 


x  =  — 


2  v2  cos2  a 
9 


y> 


382  INTEGRAL  CALCULUS 

or 

/        v2  sin  a  cos  a\2  2  v2  cos2  at        t>2sin2a\         ,AS 

and  comparing  this  with  the  equation  of  a  parabola, 

(x-h)2=  -2p(y-k), 
it  is  seen  that : 

v  sin  a.  cos  a. 

the  abscissa  of  the  vertex  = ;  (5) 

9 

v2  sin2  at 

the  ordinate  of  the  vertex  =  — = ;  (6) 

*9 

,,     ,   ,  2  v2  cos2  a  fP9S 

the  latus  rectum  = (7) 

9 

By  transferring  the  origin  to  the  vertex,  (4)  becomes 

*2=-^f%,  (8) 

which  is  the  equation  of  a  parabola  with  its  axis  vertical  and 
the  vertex  the  highest  point  of  the  curve. 

The  distance  between  the  point  of  projection  and  the  point 
where  the  projectile  strikes  the  horizontal  plane,  called  the 
Range,  is 

^D  v2  sin  2  a  rtv. 

OB  =  x  = ,  (9) 

9 

when  y  =  0,  from  (3),  which  is  evident  geometrically,  since 

OB  =  2  OC;  that  is,  the  rangt  is  equal  tojtwice  the  abscissa 

of  the  vertex. 

It  follows  from  (9)  that  the  range  is  greatest  for  a  given 

velocity  of  projection,  when  a  =  45°,  in  which   ease  the 

v2 
range  =  - .     It.  appears  from  (9)  that  the  range  is  the  same 
1/ 

for  the  complement  of  a  as  for  a.     The  greatest  height  CA 
is  given  by  (6)  which,  when  a  =  45°,  becomes  v2/4:g. 
The  height  of  the  directrix, 

CD  =  CA  +  AD  =  "-^  +  i  ?i^!^  =  |L 
20  4        g  2g 


MOTION   OF  PROJECTILE  IN   RESISTING   MEDIUM       383 

Hence,  when  a  =  45°  the  focus  of  the  parabola  is  in  the 
horizontal  line  through  the  point  of  projection,  for  then 
CA  =  \  CD. 
To  find  the  velocity  V  at  any  point  of  the  path,  from  (1), 

=  v2  cos2  a  +  (v2  sin2  a  —  2  v  sin  agt  +  g2t2) 

=  v2-2gy,    or     JJ  -  f  -  y  =  MS  -  MP  =  PS. 

V2 
Since  —  is  the  height  through  which  a  particle  must  fall 

from  rest  to  acquire  a  velocity  V,  it  follows  that  the  velocity 

at  any  point  P  on  the  curve  is  that  which  the  particle 

falling  freely  through  the  vertical  height  SP  would  acquire; 

that  is,  in  falling  from  the  directrix  to  the  curve;   and  the 

velocity  of  projection  at  0  is  that  which  the  particle  would 

acquire  in  falling  freely  through  the  height  CD. 

For  the  time  of  flight,  put  y  =  0  in  (3)  and  solve  for 

2  v2  sin  a  cos  a       ,  .  ,     ,.   .n    ,  ,  .  ,.  . 

x  = ,  which  divided  by  v  cos  a  gives,  time  of 

flight  = ;   or  in  (2)  put  y  =  0  and  solve  for  t,  giving 

r\         j     4       2  y  sin  a         ,    . 

t  =  0    and    t  =  ,  as  before. 

9 
197.  Motion  of  Projectile  in  Resisting  Medium.  —  If 
the  resistance  of  the  air  is  taken  to  vary  as  the  square  of  the 
velocity  and  the  angle  of  projection  is  very  small,  the  pro- 
jectile rising  but  a  very  little  above  the  horizontal,  the 
equation  of  the  trajectory  above  the  horizontal  line  can  be 
found.     Thus  the  equation 

gx2  gkx3 

"  ~"  2  v2  cos2  a       3v2  cos2  a 

may  be  derived  under  such  conditions;  where  the  first  two 


384  INTEGRAL  CALCULUS 

terms  represent  the  trajectory  neglecting  air  resistance,  as 
found  in  (3),  Art.  196. 

For  the  high  velocities  of  cannon-balls  the  trajectory  is 
found  to  be  very  different  from  the  parabolic  path  and  the 
range  much  less  than  that  deduced  for  it. 

Experiments  show  that  the  angle  of  projection  for  greatest 
range  is  about  34°,  rather  than  45°,  as  deduced  for  the  para- 
bolic path. 

The  simplest  formula  for  making  out  a  range  table  is 
Helie's: 


gx2     / 1    .  kx\ 
y  =  x  tan  a-     *        I— »  H )» 

"2  a  \Vo2       Vo  J 


2  cos2 


where  k  =  0.0000000458  -,  d  being  the  diameter  of  the 

projectile  in  inches,  and  w  its  weight  in  pounds. 

In  addition  to  the  resistance  of  the  air,  allowance  has  to 
be  made  in  firing  for  the  drift,  that  is,  the  tendency  for  most 
projectiles  to  bear  to  the  right  upon  leaving  the  gun,  due  to 
the  right-handed  rotation  given  to  the  projectile. 


CHAPTER   VIII. 
INFINITE  SERIES.     INTEGRATION  BY  SERIES. 

198.  Infinite  Series.  —  When  a  series  consists  of  a 
succession  of  terms  whose  values  are  fixed  by  some  law  and 
the  number  of  its  terms  is  unlimited,  it  is  an  infinite  series. 

Let  Ui,  M2,  W3,  .  .  .  be  an  infinite  succession  of  such  values 
and  let  the  sum  of  the  first  n  of  these  values  be  denoted  by 
Sn,  that  is, 

Sn  =  Ml  +  Uz  +    '    •    •     +  Un.  (1) 

When  n  becomes  infinite,  then  the  infinite  series  is 

Mi  +  Ui  +  Us  +    •   •   •   .  (2) 

If  Sn  has  a  definite  limit  as  n  becomes  infinite,  that  limit 
is  called  the  value  of  the  infinite  series,  and  the  series  is  said 
to  be  convergent. 

If  Sn  has  no  definite  limit,  either  oscillating  between  two 

finite  values  or  increasing  in  value  beyond  any  finite  value, 

the  series  is  said  to  be  divergent.     Thus  the  geometrical  series, 

a  +  ar  +  ar*  +  ar*+  •  •  •  (3) 

is  convergent  only  when  r  is  between  —  1  and  +1,  for 

Sn  =  a  +  ar-\-ar2-\-  •  •  •  +arn~l= — -  =  - , 

1  —  r          1  —  r      1  —  r 

and 

lim  Sn  =  lim  a  (*  ~  rn)  =  -?—,     when   -1<  r  <  1. 

n=oo  n=ao         L  r  1  T 

ail  —  rn) 

Since     *Sn=3C  =  — ^ =  x,     when   —  1  >  r  =  1, 

1  —  r 

and  oscillates  between  0  and  a  when  r  =  —1,  the  series 
for  those  values  of  r  is  divergent,  *S^=X  having  no  definite 
limit. 

385 


386  INTEGRAL  CALCULUS 

The  terms  of  a  series  may  be  functions  of  some  variable  x; 
then  the  series  is  said  to  converge  for  any  particular  value 
of  this  variable,  say  x  =  a,  when,  if  x  is  replaced  by  a  in  each 
term,  lim  Sn  (a)  exists.     When  the  corresponding  limits  exist 

n=oo 

for  all  values  of  x  in  a  certain  interval,  say  from  x  =  a  —  h 
to  x  =  a  +  h,  the  series  is  said  to  converge  throughout  the 
interval  and  to  define  a  function  in  the  interval. 
Example  1.  —  Let  the  given  series  be 

1  +  x  +  x2+  x3  +  •  •  •  +  x"-1  +  •  •  •  .  (1) 

\   —  %n  J 

Here    lim  Sn  (x)  =  lim  -— —  =     _     ,  for    —  1  <  x  <  1 ; 

n=oo  n=»     -L         X  1         X 

and  within  the  interval  x  =  —  1  to  x  =  +1,  not  including 

the  end  values,  the  series  defines  the  function  fix)  =  z 

1  —  x 

For  the  end  values  x  =  —  1  and  x  =  -\-  \,  and  for  all  values 

beyond,  the  series  is  divergent  and  does  not  define  a  function. 

Example  2.  —  Similarly  the  given  series, 

1  -  x  +  x2  -  x3  +  •  •  •  +  (- 1)  V-1  +  •  •  •         (2) 

has  lim>Sn(V)  =  .,    ,      ,  for    —  1  <  x  <  1, 

n=  oo  -»■  *~ r"  X 

and  within  the  interval  x  =  —1  to  x  =  +1,  the  series  de- 
fines the  function  fix)  =  t— 

J  1  -f-  x 

For  the  end  values  x  =  —1  and  x  =  +1,  and  for  all 
values  beyond,  the  series  is  divergent  and  does  not  define  a 
function. 

199.   Power  Series.  —  An  infinite  series  of  the  form 

a0  -f  aix  +  a2x2  +  a3:r3  +  •  •  •  +  anxn  +  •  •  ■  ,  (1) 
where  a0,  ai  .  .  .  ,  an,  etc.,  are  constants,  is  called  a  power 
senes  in  x.     One  of  the  form 

a0  +  ai  0  —  a)  +  a2  (x  —  a)2  +  a3  Or  —  a)3  +  •  •  • 
-r-M*-  a)»+  •  •  •  (2) 

is  called  a  power  series  in  (x  —  a). 


POWER  SERIES  387 

The  series  (1)  and  (2)  of  the  preceding  examples  (Art.  198), 

are  power  series  in  x,  representing  the  functions and 

1  —  x 

T—r —  for  certain  values  of  x.     Such  series  are  of  importance 

because  of  the  frequency  with  which  they  'occur  and  the 
special  properties  which  they  possess. 

For  instance,  the  sum  of  a  few  terms  of  an  infinite  series 
representing  some  function  may  be  a  very  close  approxima- 
tion to  the  value  of  the  function.  Thus,  if  in  the  series  (1), 
(Ex.  1,  Art.  198),  x  =  §,  the  well-known  series  converging 
to  the  value  2  results: 

1^=2=1+1  +  1  +  1+...+^+....     (3) 

If  the  terms  in  (1)  (Ex.  1,  Art.  198)  after  xk~l  be  neg- 
lected, the  error  would  be, 

.&+: 


xk  _|_  xk+*  _|_     .    .    .     _|_  xn  _j_ 


1 


This  error,  ——— ,  would  be  very  small  compared  with  the 
j.       x 

value  of  the  function,  ; ,  and  would  decrease  as  k  was 

1  —  x 

increased;  that  is,  a  closer  and  closer  approximation  would 
be  made  to  the  value  of  the  function  the  greater  the  number 
of  the  terms  retained.  For  the  particular  value  x  =  },  it 
may  be  noticed  that  the  error  made  in  stopping  with  any 
term  is  exactly  the  value  of  that  term. 

For  smaller  values  of  £  a  very  much  closer  approximation 
would  be  made  even  when  only  a  few  of  the  terms  are  taken. 
This  method  of  approximation  is  practically  useful  when  the 
exact  value  of  the  function  is  unknown  or  does  not  admit  of 
exact  numerical  expression,  for  examples,  the  numbers  e  and 
7r,  the  logarithms  of  numbers,  and  the  trigonometric  func- 
tions of  angles  in  general.  In  Articles  to  follow  power  series 
for  such  functions  will  be  given. 


388  INTEGRAL  CALCULUS 

200.  Absolutely  Convergent  Series.  —  A  series  the  abso- 
lute values  of  whose  terms  form  a  convergent  series  is  said  to 
be  absolutely  convergent ;  other  convergent  series  are  said  to 
be  conditionally  convergent.     For  example,  the  series 

1  -  I  +  i  -  i  +  '  •  •     (Ex.  2,  Art.  198),  (1) 

is  an  absolutely  convergent  series,  since  it  converges  when 
all  terms  are  given  the  positive  sign,  as  (3),  Art.  199.  On 
the  other  hand,  the  series 

i-i  +  i-i  +  1...  (2) 

is  conditionally  convergent,  since  the  series 

resulting  from  making  all  terms  positive,  is  divergent. 
The  series  (3)  may  be  seen  to  be  divergent  in  the  form, 


2  '  V3 


+  S+d^+---+2F=-i)+---> 

for  the  sum  of  the  terms  in  each  of  the  parentheses  is  greater 
than  |,  and  as  the  number  m  of  such  groups  that  can  be 
formed  in  the  given  series  is  unlimited, 

Sn=oc  >  (m  X  i)  =  oo. 
This  result  reveals  the  important  fact  that  while  the  defini- 
tion of  an  infinite  convergent  series,  requiring  Sn=ao  to  have 
a  definite  limit   (Art.  198),  makes  lim  un  =  Q  a  necessary 

n=oo 

condition,  that  condition  is  not  sufficient  to  insure  conver- 
gency.  In  other  words,  a  series  is  not  necessarily  convergent 
when  the  terms  themselves  decrease  and  approach  zero,  as 
the  number  of  terms  increases  without  limit.  For  the  series 
(3),  the  condition  is  fulfilled;  the  nth  term  approaches  zero 
as  a  limit,  as  n  increases  without  limit,  and  yet  the  sum  of  the 
first  n  terms  has  no  limit  and  the  series  is  divergent. 


ABSOLUTELY  CONVERGENT  SERIES  389 

When,  however,  the  terms  of  the  decreasing  series  are 
alternately  positive  and  negative  the  condition  is  sufficient 
to  prove  convergency.  Thus  the  series  (2)  being  such  a 
series  is  convergent,  though  not  absolutely  so.  This  series 
may  be  put  in  the  forms, 

(i  -  i)  +  (i  -  i)  +  tt  -  i)  +  •  •  • , 

01'  2  "h  T1 2  +  3"V  +    *    *    '    , 

i  -  (i  -  i)  -  (i  -  «  •  .  .  , 

or  1  -  i  -  2]o  -   •  '  *  , 

where  it  can  be  seen  that  the  sum  of  n  terms  of  the  series  is 
greater  than  \  and  less  than  1.  It  will  be  shown  further  on 
that  the  limit  of  the  sum  is  log  2  =;0.69.  .  .  (Ex.  1,  Art.  203). 
While  it  is  thus  seen  that  the  series  (2)  is  conditionally 
convergent,  the  following  series  is  absolutely  convergent : 

1  -  I      I      I  4- 1  -   .  (A\ 

22      32      42      52  ' 

since  1  +  ^  +  ^  +  ^  +  -2  +  •  •  •  (5) 

is  convergent,  as  may  be  shown  by  comparison  with  the  series 

2      22      23      24  ' 

or  1 +  *  +  }  +  *+••-,     (3),  Art.  199, 

known  to  be  convergent. 

The  series  (5)  is  the  more  general  series 

when  p  =  2;    and  the  series  (3)  of  this  Article,  called  the 
harmonic  series,  is  the  series  (6)  when  p  =  1. 

This  series  (6)  is,  therefore,  divergent  for  p  =  1;  for 
p  <  1,  every  term  after  the  first  is  greater  than  the  corre- 
sponding term  of  the  series  (3),  hence  (6)  is  divergent  in  this 
case  also. 


390  INTEGRAL  CALCULUS 

For  p  >  1,  compare 

I? T  ^      3"/      V4»      5"  T6»      1") 

+  (P+--  -+lfc)+'--     (6) 
with 

i+zi+iwi+i+i+ri 

1p  '  y2p      2P/      \4P      4P      4P      4P/ 

'  +(£+••.•+£)+•■■  m 

There  is  the  same  number  of  terms  in  the  corresponding 
groups  of  the  two  series  and  the  sum  of  the  terms  in  those  of 

(6)  is  in  each  group  less  than  the  sum  in  those  of  (7).     Now 

(7)  may  be  put  in  the  form 

1+2,1,8 

2 

a  geometrical  series,  whose  ratio,  — ,  is  less  than  unity.    Hence 

by  (3)  of  Art.  198,  (7)  is  convergent  and  consequently  (6) 
is  convergent. 

The  series  (6)  and  the  geometrical  series  are  useful  as 
standard  series,  with  which  others  may  often  be  compared 
to  test  convergency. 

201.  Tests  for  Convergency.  —  It  has  been  seen  that  a 
conditional  test  is  that  in  every  convergent  series  the  nth  term 
must  approach  zero  as  a  limit,  as  n  is  increased  without  limit. 

This  condition  is  involved  in  and  may  be  deduced  from 
the  definition  (Art.  198),  that  the  infinite  series 

Ui  +  u2  +  m  +  •  •  •  (1) 

is  convergent,  if  lim  Sn  exists. 

n=co 

For  since 

Sn  =  Sn-i  +  un,  lim  Sn  =  lim  £n_i  +  lim  un  =  lim  Sn-i, 

#  =  oo  n=oo  n=oo  n=oo 

if  lim  Sn  exists ;     .*.     lim  un  =  0.     The  converse  is  not  true. 


TESTS  FOR  CONVERGENCY  391 

In  the  same  way  it  may  be  shown  that  the  Remainder  after 
n  terms  of  a  convergent  series  must  approach  zero  as  a  limit 
as  n  becomes  infinite;  thus,  letting  Rn  denote  the  series  of 
terms  after  the  nth, 

Rn   =   Uu+l  +  Un+2  +  Un+3  +     '     '     '     . 

Now  if  S  denotes  the  value  of  the  series  (1),  it  is  the  limit  of 
the  sum  of  the  terms  in  the  series;  that  is, 

S  =  lim  (Sn  +  Rn)  =\miSn; 
:.     lim  Rn  =  0. 

For  example,  in  (3),  Art.  199,  Rn  =  (i)""1;    .*•    lim  Rn '  =  0. 

It  was  seen  in  (3),  Art.  200,  that  although  the  nth  term 
approaches  zero,  the  series  is  not  convergent,  hence  the 
remainder  after  n  terms  does  not  approach  zero  as  a  limit  in 
that  series,  as  it  is  a  divergent  series. 

For  convergency,  lim  Rn  =  0  is  a  decisive  test;  Km  un  =  0 

n=oo  71=  x 

is  a  decisive  test  only  when  the  terms  alternately  have  dif- 
ferent signs;  it  is  a  conditional  test  when  all  the  terms  have 
the  same  sign.     For  divergency,  lim  un  not  equal  to  zero  is  a 

n=x 

decisive  test.     Hence,  when  lim  un  =  0,   unless  the  terms 

alternately  have  different  signs,  the  test  is  indecisive. 

The  Comparison  Test.  —  It  may  often  be  determined 
whether  a  given  series  of  positive  terms  is  convergent  or 
divergent,  by  comparing  its  terms  with  those  of  another 
series  known  to  be  convergent  or  divergent.  The  method 
of  applying  this  test  and  the  standard  series  useful  for  com- 
parison have  been  given  in  the  preceding  Article  200.  This 
test  is  often  available  when  other  tests  fail  to  be  decisive. 

The  Ratio  Test.  —  A  given  series 

Wl  +  "2  +«•+•'•+  Un~l  +  Un  +    •    •    • 

u 
is  convergent  or  divergent,  according  as  lim  ■ — —  is  less  or  greater 

n=<x>  Un—l 

than  1. 


392  INTEGRAL  CALCULUS 

This  test  applies  when  some  of  the  terms  of  the  series  are 
negative  as  well  as  when  they  are  all  positive.     It  is  no  test, 

however,  when  lim  — —  =  1 ;  in  that  case  other  tests  must 

n=oo  Un—1 

be  applied. 

For  example,  let  the  given  series  be 

Here        lim  — -  =  lim  ( — .  /  -, r^r )  —  nm  -  =  0 

n  =  ooUn-i        „=oo\w!/     (n—\)\J         n  =  ooTl 

for  any  finite  value  of  x.     Hence  the  series  (2)  converges  for 
all  finite  values  of  x. 

This  series  (2),  as  given  in  Art.  36,  is  the  expansion  of  e*; 
and,  when  x  =  1,  the  limit  of  the  series  is  the  number  e,  in 
Art.  34  (see  Ex.  5,  Art.  215,  also).  The  ratio  test  is  found 
to  be  true  by  comparing  the  given  series  with  the  geometrical 

series;  hence,  when  lim  — —  =  1  and  the  test  fails  to  be  de- 

n=oc  1l>n—i 

cisive,  recourse  is  to  the  comparison  test,  as  shown  in  Art. 
200,  for  the  harmonic  series 

1+2+3+  n^ 


EXERCISE   XLII. 

1.   Find  whether  the  following  series  are  convergent  or  divergent: 

(i)  l  +*  +  *  +!+••• . 

(2)  I+J  +  I  +  J+---. 

(3)  F2  +  2^3  +  3^1+4^5+  '  '  *  ' 

(4)  1+I  +  I  +  I  +  I+.... 

(5)  ^f!  +  f!  +  f!+---- 

02  Q3  44 

(6)  l  +  2!  +  3]  +  4!+'--- 


CONVERGENCE  OF  POWER  SERIES 


393 


wl+l+T0+ 


:+5 


n2  +  l 


(9) 


1        +-^+        * 


1  +  VI      1  +  V2       1  +  V3 


(10)    l  +  ^  +  |  +  ^  +  ^  + 

2.   Examine  the  following  series  for  convergency: 

(1)  logf-logf  +log|-logf  +  •  •  •  . 

(2)  sec  =  —  sec  -.  +  sec  =  —  sec  3  +  •  •  •  . 

6  4  5  o 

(3)  sin2£  +  sin2^  +  sin2?  +  sin2?  +  •  •  •  . 

2,  6  4  o 

202.  Convergence  of  Power  Series.  —  A  power  series  in 
x  may  converge  for  all  values  of  x,  as  in  (2),  example  of  the 
Ratio  Test ;  but  generally  it  will  converge  for  certain  values 
of  x  and  diverge  for  others,  as  in  Examples  1  and  2,  Art.  198. 

Applying  the  ratio  test  to  the  power  series, 

a0  +  aix  +  (hx2  +  asx3  +  •  •  •  +  anxn  +  •  •  •  .     (1)  (Art,  199.) 


un-\      an-i ' 

im 

n=ao 

Un 

=  lim    ■ 

71=00  1  &n— 1 

= 

\x\] 
1 

im 

an 

The  series  (1)  is  convergent  or  divergent  according  as 

|  x  |  lim 

n=oo 

an 

dn-1 

<  1,     or     |  x  |  lim 

7i  =  SC 

an 

fln-l 

>i; 

that  is,  according  as 

|  x  \  <  lim 

Gn-1 
Cln 

,     or     1  3 1  >  lim   -^ 

The  case  |  x  |  = 

iim 

an-i 
an 

is  undecidec 

lb 

y  thi 

3  tes 

3t. 

When,  however,  a  power  series  is  convergent  for  any  value 
of  x,  say  a,  it  is  absolutely  convergent  for  all  values  of  x  such 
that  |  x  |  <  |  a  |. 

For  example,  given  the  series 
l+2z  +  3x2  +  4z3+  •  •  •  nxn~l  +  (n  +  1)  x"  H ;       (2) 


394  INTEGRAL  CALCULUS 

.  an-i  n  ,.         n  ,.         1  ■ 

here  =  — r-r ,     lim  — —  =  lim r  =  1 ; 

an       w  +  1      n=00n  +  l      n=a0  ^   ,   1 

n 
hence  (2)  is  convergent  or  divergent,  according  as 
|z|<l     or     |z"|>l; 

that  is,  (2)  is  convergent  when  —  1  <  x  <  1,  and  the  interval 
from  —1  to  +1  is  the  interval  of  convergence.  Here  the 
interval  does  not  include  the  end  values,.  (2)  being  divergent 
when  |  x  |  ==  1. 

203.  Integration  and  Differentiation  of  Series.  —  A 
power  series  has  the  important  property  that,  when  the 
variable  of  the  function  is  restricted  to  the  interval  of  con- 
vergence, it  is  possible  to  get  the  integral  or  the  derivative  of 
the  function  by  integrating  or  differentiating  term  by  term 
the  series  which  defines  the  function.  Hence,  if  /  (x)  is 
defined  by  the  power  series, 

f(x)  =  a0  +  aw  +  a2x2  -f  •  •  •  +  cinXn  +  •  •  •  ;       (1) 
then 

Jf(x)dx=  f     a0dx+  I    xaixdx-\-  •  •  •  +  /     anxndx+  •  •  • , 
X0  *J Xq  t/x0  J  Xq 

and 

df(x)  =  d(a0)       d  (aix)  d  (anxn) 

dx  dx    ■*"     dx      >   '     '  "*"  ~llx       h  '  '  '  ' 

when  the  restriction  necessary  to  insure  convergence  is  placed 
upon  the  value  of  x. 

Example  1.  —  For  —  1  <  x  <  1, 

^^  =  1  -  x  +  x2  -  Xs  +  •  •  •  .  (Ex.  2,  Art.  198.)  (2) 
Hence, 

Jn —  =    /    dx  —    I    xdx+  \    x2  dx  —  I    xz  dx  +  •  •  •  , 
o   1  +  x       Jo  Jo  Jo.  Jo 

thatis,       log(l+a:)  =  z-*!2  +  f-  J+  •  •  •  .  (1) 


INTEGRATION   AND   DIFFERENTIATION  OF  SERIES     395 

This  is  the  logarithmic  series  with  base  e,  and  is  true  for 
—  1  <  x  =  1;  that  is,  it  is  true  for  values  of  x  within  the 
original  interval  of  convergence,  including  the  end  value  1; 
but  for  the  other  end  value  —1,  it  decreases  without  limit. 
On  putting  x  =  1  in  (1), 

log2=l-J  +  i-i+---=  0.69  ....     (See  Art.  200.) 

On  putting  x  =  —  1  in  (1), 
log0=  -  (l  +  i  +  i  +  i+  •  •  ■  )  =  -».     (See  Art.  200.) 

In  the  same  way,  for  —  1  <  x  <  1,  the  integration  of 

1 


— —  =  1  +  x  +  X2  +  x*  +   ■ 
1  —  3! 

•  •      (Ex.  1,  Art.  198)     (1) 

gives      log  (1  —  x)  =  —  x  —  \  x2  - 

.  ir3_ix4 }      (2) 

which  may  be  gotten  by  putting 

—x  for  x  in  (1). 

This  logarithmic  series  is  true  for  —  1  =  x  <  1;  that  is, 
it  is  true  for  values  of  x  within  the  interval  of  convergence, 
including  the  end  value  —1;  but  for  the  other  end  value  1, 
it  decreases  without  limit. 

For  x  =  —  1,  (2)  gives  log  2  as  above,  and  for  x  =  1, 
log  0  as  above.  Hence  by  neither  series  can  the  logarithm  of 
a  number  greater  than  2  be  found.  Bv  a  combination  of 
the  two  series  the  logarithm  of  any  number  can  be  found. 

By  subtracting  (2)  from  (1), 

1  +  x 

log  (l  +  x}  -  log  (i  -  x)  =  log  yzt^ 

=  2(z+!  +  !°+-  •  •),   for  |  x  |  <  1.      (3) 

For  x  =  \, 
logi±|  =  log2  =  2g+3^+^+^+.  •  .)  =  0.6931  ..  .. 

For  x  =  ^ 
^[±1  =  ^3  =  2(1+3^+^+.^+...) 


1.0986  . 


396  INTEGRAL  CALCULUS 

This  series   (3)   converges  very  much  more  rapidly  for 

values  of  x  less  than  1  than  the  series  (1),  which  converges 

so  slowly  that  100  terms  give  only  the  first  two   decimals 

correctly  for  the  log  2,  while  (3)  gives  four  decimals  correctly 

taking  only  four  terms  of  the  series.     Any  number  may  be 

1  +  x 

put  in  the  form ,  but  it  is  necessary  to  calculate  directly 

j.      x 

the  logarithms  of  the  prime  numbers  2,  3,  5,  7  only,  as  the 

others  can  be  expressed  in  terms  of  these.     Thus, 

1  +  -  5  5 

log  Yz^i  =  loS  3  >  and  then  lo§  5  =  log  -  +  log  3; 

and  again, 

I  _i_  i  7  7 

log  - f-  =  log  -,  and  then  log  7  =  log  -  +  log  5. 

1  —  6  o  o 

To  get  the  common  logarithms  whose  base  is  10,  multiply 
these  natural  logarithms  by  0.4343  .  .  .  ,  the  modulus  of 
the  common  system.     (See  Art.  38  and  Ex.  6,  Art.  215.) 

Example  2.  —  For  —  1  <  x  <  1, 

— —  =  1  +  x  +  x2  +  xs  +  •  •  •  .     (Ex.  1,  Art.  198.)     (1) 
1  —  x 

By  differentiation, 

1+2z+3x2+4.t3H .  (See  (2),  Art.  202.)     (2) 


(1  -  x)2 

By  differentiation  again 


=  £(1. 2  +  2-33  +  3. 4s*+  •  •  •  ).        (3) 


(1  -  xf      2 
Hence,  the  general  series, 


1  t\         \  »      1    1  ,  m  (m  +  1)    9 

=  (1  —  x)~m  =  1  +  mx  H ^-^tt1 — -x2 


(1  -  x)m      v  '  '  '  2! 

+  w(nt  +  l)(w  +  2)g,+  >>  (4) 


INTEGRATION  AND   DIFFERENTIATION  OF   SERIES     397 

Example  3.  —  For  -1  <  x  <  1,  by  (4)  of  Ex.  2,  or  by 
division, 

rq^=i-z2+*4 •  (i) 

Hence,       /    — - — r  =  /    dx  —    /    x2dx  +    /    x4dx—  •  •  •  , 
Jo   1  +  x?    .  Jo  Jo  Jo 

that  is, 

arc  tan  x  =  x  —  ^-  +  —  —  •••.  (2) 

o        o 

This  is  Gregory's  series,  named  after  its  discoverer,  James 
Gregory. 

Although  series  (1)  oscillates  when  x  =  1,  series  (2)  is 
convergent  and  defines  arc  tan  x  even  when  x  =  1.  On 
putting  x  =  1, 

arctanl  =  j=l-g  +  --7+  •  •  •  ; 

,  ^(,-1+1-1+...). 

While  the  value  of  w  may  be  found  approximately  from  this 
series,  the  series  converges  so  slowly  that  it  is  better  to  use 
other  more  rapidly  convergent  series,  such  as, 


and 


7T  1  1 

-  =  4  arc  tan  -  —  arc  tan  ^r—  (Machin's  Series), 

4  O  Zoo 


-  —  arc  tan  -  +  arc  tan  -  •     (Euler's  Series.) 


Example  4.  —  For  - 1  <  x  <  1,  by  (4)  of  Ex.  2, 

=  (l-x2)-*  =  l+^x2+--ix*+7r-r-;xs  + 


Vl-x2  2      '  2-4      '  2-4-6 

hence, 

C *     dx  .  ,  1    x3  ,  1  •  3   x5  .  1  -  3  •  5    x1  , 

J0VT^=arCSmX  =  a:+2-3+2T4'5+^4T6'7+"-- 

This  series,  due  to  Newton  and  used  by  him  to  compute  the 
value  of  T  approximately,  converges  rapidly  for  x  <  1. 


398  INTEGRAL  CALCULUS 

When  x  =  J,  this  series  gives 

.    1     7T     1  ,     *  1  1-3  1-3-5 

arc  sin  -  =  -  =  -+0   o    »a+r>    ,    r    os+ 


2     6     2'2-3-23,2-4.5.25,2.4.6.7-271 

To  ten  places,       tt  =  3.1415926536  .... 

By  means  of  series  the  value  of  tt  has  been  carried  to  700 
decimal  places. 


J  fa  x  fa 

(a2  —  e2x2) 2     .  This 

o  v  a2  —  x2 


can- 


not be  integrated  directly,  but  on  expanding  (a2  —  e2x2)*  by 
the  binomial  theorem  the  terms  of  the  resulting  convergent 
series  can  be  integrated  separately.     Thus, 

i  P  X  P  T 

(a2  —  e2x2)*  =  a  —  t—  —  ^-z  —  •  •  •  ,    where  e  <  1,     (1) 
2  a       8  a3 


/    2  o    2\i  t*»t/ 

"  e  X  j   Va2  -  z2     (See  Ex.  1,  Exercise  XXV.) 


•  x 

J  "<>>      dx  e2    P*    x2  dx         e4     Ca  _^dx_ 

o  Va2-x2     2aJ0  Va?-x2     8a3 J0  Va2-f       v- 

=  Wi  _  6_! _  /L3Yi4 -  /i-3-5yc«  \       m 

"  2  V        22      \2-4/  3       V2-4.6/  5  '/  w 

=  a.T   (1  -e2sin20)*d0; 

T?/i      i  •  •  ««     i  a  •  i*  n  ^ /See  Ex.  6,  Exer-\ 

=  a\    (l-|e2sin20-ie4sin40-  •  •  •)  d0  •     vvn 

Jo  V     cise  XXII.      / 

"  2(1-(2Jg2-(2T4J3-(2-T4T6J5 }      (2) 


When  a;  = 

IT 

a  sin  0, 

-  e2z2)* 

—  e2  sin 

dx 

Vo2- 
2 6)*  dd 

.r2 

INTEGRATION  AND  DIFFERENTIATION  OF  SERIES     399 

Example  6.  —  Given 

P    ,        2adX  (See  Art.  236.) 

Jo  V2g{h-x)(2ax-x2) 

This  does  not  admit  of  direct  integration,  but  on  expanding 
it  into  a  power  series  in  x/2a  the  integral  can  be  evaluated 
approximately.     Thus, 

fh  2adx  =     i~a  fh       dx        L  _   x\-% 

Jo  V2g(h-x)(2ax-x2)      VaJo   Vhx  -  x2  \        2a/ 
by  (4)  of  Ex.  2, 

dx 


m^m^Mh-i 


Vhx  —  x2 
by  integrating, 

=^MI)'2V(H)'fe)' 

<mm'+  ■  ■ } 

When  h  is  small  in  comparison  with  a,  all  terms  containing 

^—  may  be  neglected,  and  the  approximate  value  is  -k  i /-. 
2  a  y  g 

If  the  given  integral  is  put  in  the  form 

2  V  -  f^  (1  ~  k2  sin2  4>Th  dcf>;     [Art.  236 
T  Q  Jo 

then, 

,—      «■ 

2V-  P(l -fc2  sin2  </>)"*  o> 
y  QJ«  _      r  [by  (4)  of  Ex.  2.] 

=  2y^^l+i/c2sin20  +  |.|fc4sin40+  .  .  .  )  0>, 
Ity  integrating, 


400  INTEGRAL  CALCULUS 

When  k  is  small,  the  approximate  value  of  the  integral  is 

i  la 
again  -k  y  -• 

Note.  —  The  integral  forms  in  Examples  5  and  6  are  called 
elliptic  integrals. 

The  forms     F  -7=M==    and     f   Vl  -  e2  sin2  0  d4> 
Jo   Vl  -  /c2sin20  Jo 

are  known  respectively  as  "  elliptic  integrals  of  the  first  and 
the  second  kind." 

a 
In  the  first  kind  k  =  sin  ^ ;  and,  in  Ex.  6,  a  is  the  angle 

each  side  of  the  vertical  through  which  a  pendulum  of  length 
a  vibrates,  the  approximate  time  of  a  vibration  being  t  y  - , 

as  found.     (See  Art.  236.)     Tables  give  values  for  varying 
values  of  a. 

In  Ex.  5,  e  is  the  eccentricity  of  an  ellipse  and  6  is  the 
complement  of  the  eccentric  angle.  By  taking  a  few  terms 
of  the  final  series,  when  e  is  small,  an  approximate  value  of  an 
elliptic  arc  of  a  quadrant's  length  is  obtained.  When  e  =  0 
the  result  is  the  length  of  a  quadrant  of  the  circumference  of 
a  circle. 


CHAPTER   IX. 

TAYLOR'S  THEOREM  —  EXPANSION  OF  FUNC- 
TIONS.    INDETERMINATE   FORMS. 

204.   Law   of   the    Mean.  —  The   mean   value   of  /  (x) 
between  the  values  /  (a)  and  /  (b)  is,  by  Art.  133, 


/   f(x)dx 
f(c)- 


b  —  a 

where  c  is  some  value  between  a  and  b. 

If  the  function  of  x  is  <j>  (x)  =  f  (x)  and  xx  is  some  constant 
value  between  a  and  x,  then 

x  —  a  x  —  a 

or  f(x)=f(a)+f(x1)(x-a),  (1) 

the  Law  of  the  Mean,  or  Theorem  of  Mean  Value. 

If  the  curve  in  the  figure  be  the  graph  of  y  =  f(x) ;  then 
the  ordinate  at  Pi  will  be  f'(xi),  the  mean  value  of  f'(x) 
between/' (a)  at  Pa  and/'(z)  at  any  value  x,  the  integral  being 
represented  by  the  area  under  the  curve  from  x  =  a  to  x  =  x. 

If  the  curve  is  y  =  f  (x) ,  it  may  be  seen  that  there  must  be 
at  least  one  point  Pi  between  the  points  (a,  /  (a))  and  (x, 
f  (x))  at  which  the  slope  of  the  tangent  is  equal  to  the  slope 
of  the  secant  through  those  points;  that  is 

r(gj-/w-/w-|g,' 

J  v    '  x-  a  Ax 

and  hence  (1). 

401- 


402  INTEGRAL  CALCULUS 

This  may  be  put  in  the  form, 

which  may  be  used  to  determine  increments  approximately, 
and  is  the  Theorem  of  Finite  Differences. 


The  theorem  may  be  extended  so  as  to  express  in  terms  of 
the  second  derivative  the  difference  made  in  using  the  first 
derivative  at  x  =  a  in  place  of  its  value  at  x  =  X\.  Thus, 
if  the  function  of  x  is  <j>  (x)  =  f"  (x),  and  x2  is  some  constant 
value  between  a  and  x,  then, 

J  J"  (x)  dx 
_a f[  (x)  -  /'  (a)      /by  mean  value, \ 

J    KX2)  ~        x-a  x-a       '    \      Art.  133,       ) 

or  f'(x)=f'(a)+f"(x2)(x-a). 

Integrating  this  equation  between  the  limits  x  =  a  and 
x  =  x,  f  (a)  and  /"  (x2)  being  constants,  gives 

/(*)  =f(a)  +f  (a)  (x  -  a)  +/"  fe)  ii^!,        (2) 
a  second  Theorem  of  Mean  Value,  or  the  Law  of  the  Mean. 


OTHER  FORMS  OF  THE  LAW  OF  THE  MEAN   403 

If  the  tangent  at  Pa  meet  the  ordinate  MP  produced  at  R, 
then,       MR=f  (a)  +  /'  (a)  (x  -  a) ;     MP  =  f  (x) , 
and,  therefore,  both  in  sign  and  in  magnitude, 

RP  =  MP  -  MR  =  /"  ta)  (*  ~  a)2- 

Here  the  deviation  of  the  curve  at  P  is  below  the  tangent  at 
Pa,  f"  fe)  being  negative,  and,  measured  along  the  line  of 

(x  _  a)2 

the  ordinate  MP,  is  equal  to  /"  (x2)  - — s-^-.     When  the 

curve  is  above  the  tangent, }"  (x2)  will  be  positive  and  RP 
upward. 

205.   Other  Forms   of  the   Law   of  the   Mean.  —  The 

theorems  (1)  and  (2)  may  be  given  in  the  following  forms. 

In  the  theorems  x,  the  symbol  for  the  argument  in  general 
has  been  used  for  any  value  of  the  argument,  a  definite  value 
but  not  constant.  Now,  if  Xi  be  any  number  between  a  and 
x,  then  xi  —  a  and  x  —  a  are  of  the  same  sign  whether  a  is 
less  or  greater  than  x;  therefore,  (xi  —  a)/ (x  —  a)  is  a  posi- 
tive proper  fraction,  6  say,  and  xi  =  a  +  6  (x  —  a)  will 
denote  any  number  between  a  and  x. 

Letting  x  =  a  -\-  h,  x  —  a  =  h;  then  theorem  (1)  will 
become 

f(a  +  h)  =f(a)+hf(a  +  dh),  (lfl) 

and  theorem  (2)  will  become 

/  (a  +  h)  =  /  (a)  +  hf  (a)  +  |/"  (a  +  eji) .  (2a) 

The  0i  of  (2a)  is  not  necessarily  the  same  as  the  6  of  (10). 
If  a  is  replaced  by  x,  the  forms  become 

f(x  +  h)  =f(x)+hf(x  +  dh),  (h) 

f(x  +  h)=f(x)  +  hf  (x)  +  |/"  (*  +  M)  •  (26) 


404  INTEGRAL  CALCULUS 

If  a  is  made  zero  and  then  x  is  put  for  ft,  the  forms  are 

f(x)=f(0)+xf(dx),  (lc) 

f(x)=f(0)+xf(0)  +  p"(Blx).  (2c) 

Example  1.  —  To  find  the  value  of  0,  if  f(x)  =  x2.     Here 
f(x)  =  2x;    /'(a  +  0ft)  =  2(a  +  0ft); 
and      (a  +  ft)2  =  a2  +  2aft  +  ft2  =  a2  +  ft  •  2  (a  +  Jft); 
also,  by  (10), 

(a  +  h)2  =/(o)  +  ft/'  (a  +  0ft)  =  a2  +  ft  •  2  (a  +  0ft); 

hence  in  this  case  6  =  \. 

To  find  at  what  point  on  the  parabola  y  =  x2,  the  tangent 
is  parallel  to  the  secant  through  the  points  where  x  =  1  and 
x  =  3.     By  theorem  (1), 

.*.     xi  =  2;  or  by  theorem  (10),  Xi  =  a  +  0ft  =  1  +  \  (2)  =  2, 
since  0  =  ■§. 

Example  2.  —  To  find  at  what  point  on  the  curve  y  = 
sin  x,  the  tangent  is  parallel  to  the  secant  through  the  points 
where  x  =  30°  and  x  =  31°.     Here 

Mt    N  sin31°-sin30°      0.51504-0.5      n0£M_„ 

/  (ft)  =  cos*  =       31o_30o       -       0.01745       =  °'86177; 

/.     *  =  cos-1  0.86177  =  30°  29'; 
hence  yx  =  sin  30°  29'  =  0.50729,     0  =  |J. 

Example  3.  —  To  show  that  sin  a;  is  less  than  a:  but  is 
greater  than  x  —  \x2. 

f(x)  =  sin  a:;    f'(x)  =  cos  a;;    /"(a;)  =  —  sin  x; 
/(0)  =  0;    f  (0)  =  1;    /"  (0!*)  =  -  sin  (0^). 
By  theorem  (lc), 

sin  a:  =  0  +  x  cos  (0a:),   <  x,  since  cos  (Ox)  <  1. 


TAYLOR'S  THEOREM  405 


By  theorem  (2C), 
sin x  =  0  +  x  —  Tr-sin  (Six)  >  x  —  — ,  since  |  sin  (Six)  \  <  1. 

Example  4.  —  To  show  that  cos  x  is  greater  than  1  —  \  x2. 

f(x)  =  cos  a:;    f'(x)  =  —  sinz;    f"(x)  =  -  cos  a;; 
/(0)  =  1;    /'  (0)  -  0;    /"  (M  =  -  cos  (dlX). 

By  theorem  (2C), 

cos  a;  =  1  —  |  z2  cos  (^io;)  >  1  —  \  x2,  since  |  cos  (fiix)  |  <  1. 

206.  Extended  Law  of  the  Mean.  —  The  law  of  the 
mean  or  the  theorem  of  mean  value  in  its  several  forms  may 
be  used  to  obtain  approximate  expressions  for  a  given 
function  in  the  neighborhood  of  a  given  point  x  =  a.  Still 
closer  approximations  may  be  obtained  from  the  law  when 
extended  in  the  form  of  a  series  arranged  according  to  ascend- 
ing powers  of  x  —  a  with  the  successive  derivatives  as  con- 
stant coefficients. 

For  values  of  x  near  to  a,  the  higher  powers  of  x  —  a  may 
then  become  negligible.  The  most  convenient  theorem  for 
this  purpose  is  the  one  which  follows. 

207.  Taylor's  Theorem.  — If  f  (x)  is  continuous,  and  has 
derivatives  through  the  nth,  in  the  neighborhood  of  a  given  -point 
x  =  a,  then,  for  any  value  of  x  in  this  neighborhood, 

f(x)=f(a)+f-^-(x-a)+f-^f(x-ay+  ■  ■  ■ 

where  X  is  some  unknown  quantity  between  a  and  x. 
The  last  term 

is  the  error  in  stopping  the  series  with  the  nth  term,  the  term 
in  {x  —  a)  n~1 ;  and  the  formula  is  of  practical  use  only  when 


406  INTEGRAL  CALCULUS 

this  error  becomes  smaller  and  smaller  as  the  number  of 
terms  is  increased. 

The  form  of  the  remainder  Rn(x)  is  seen  to  differ  from  the 
general  term  of  the  series  only  in  that  the  derivative  in  the 
coefficient  of  the  power  of  (x  —  a)  is  taken  for  x  =  X  instead 
of  for  x  =  a. 

The  simplest  proof  of  this  theorem  is  the  extension  by 
integration  of  the  law  of  the  mean  —  a  further  extension 
than  already  used  for  the  theorems  of  mean  value. 

Thus,  if  the  function  of  x  is  <£  (x)  =  /'"  (x),  and  X  is  some 
unknown  constant  value  between  a  and  x,  then, 


x 


/'"  (x)  dx  =  /'"  (X)  (x  -  a)     (by  mean  value,  Art.  133) 
=  f"  (x)  -  f"  (a)  =  f"  (X)  (x  -  a). 


Integrating  this  equation  between  the  limits  x  =  a  and 

x  —  Xj 

J1  (x)  -  f  (a)  -  f"  (a)  (x  -  a)  =  /'"  (X)  ^^, 

/"  (a)  and  /'"  (X)  being  constants. 
Integrating  again,  /'  (a)  also  being  constant,  gives 

f(x)  =  f(a)  +f  (a)  (x  -  a)  +/"  (a)  ^-^  +/'"  (X) (a 


2-3 

As  this  process  can  be  continued  to  include  the  nth  deriv- 
ative, by  induction  Taylor's  Theorem  results. 

208.  Another  Form  of  Taylor's  Theorem.  —  If  in  the 
form  (1)  x  is  put  for  a  and  (x  +  h)  for  x,  it  becomes 

six + h)  =/(x)  +/'  (x)  \+r  (x>|  +  •  •  ■ 

+/""1(a;)(S)!+/"(*  +  <Wl)S'  (2) 

where  the  last  term  is  the  remainder  after  n  terms,  and  0  is 
some  positive  proper  fraction.  In  (1),  (a  +  6  (x  —  a))  may 
be  used  in  place  of  X;  and  in  (2),  it  becomes  (x  -\-  Oh) 


MACLAURIN'S  THEOREM  407 

Another  form  of  the  remainder  called  Cauchy's  is 

i?„  (s)  =  f  (a  +  8  (x  -  a))  {x  ~  g"^1  ~,  9)*"'  • 

iVo^e.  —  Taylor's  Theorem  is  the  discovery  of  Dr.  Brook 

Taylor,  and  was  first  published  by  him  in  1715.     He  gave  it 

as  a  corollary  to  a  theorem  in  Finite  Differences  and  there  was 

no  reference  to  a  remainder.     It  was  Lagrange  who,  in  1772, 

called  attention  to  its  great  value  and  found  for  the  remainder 

xl                 •       f[a-{-d(x  —  a)],  x       .  n    ii      i- 

the  expression,  — -. —  (x  —  a)n,  since  called  by  his 

name.  It  has  become  to  be  regarded  as  the  most  important 
formula  in  the  Calculus. 

The  formula  known  as  Maclaurin's  Theorem,  after  Colin 
Maclaurin,  was  published  by  him  in  1742,  but  he  recognized 
it  as  a  special  case  of  Taylor's  Theorem.  The  two  theorems 
are  virtually  identical  as  either  can  be  deduced  from  the 
other. 

209.  Maclaurin's  Theorem.  —  If,  in  the  form  (1),  a  is 
made  0; 

+  {n-l)\X      +     n\    X'  (S) 

where  X  =  dx  is  some  unknown  quantity  between  0  and  x. 
Another  form  of  Maclaurin's  Theorem  is 

m  -/«»+q°>x  +«*-+  ••• 

/-HO)*"-1  ,  fn(ex)xn 

"*"    (w-  1)!     **       n\       '  w 

where  the  last  term  is  the  remainder  after  n  terms,  and  0  is 
some  positive  proper  fraction. 

Cauchy's  form  of  the  remainder  is 

xn(l  -  0)"-1 


Rn  (x)  =  /» (ex) 


(n-1)! 


408  INTEGRAL  CALCULUS 

210.  Expansion  of  Functions  in  Series.  —  It  has  been 
shown  in  examples  of  Chapter  VIII  on  Infinite  Series  how 
useful  it  is  to  be  able  to  represent  a  function  by  means  of  a 
series.  Apart  from  the  purpose  of  computation  such  rep- 
resentation may  be  an  aid  to  an  understanding  of  the  proper- 
ties of  functions.  Taylor's  Theorem  and  Maclaurin's 
Theorem  furnish  a  general  method  of  expanding  or  develop- 
ing any  one  of  a  numerous  class  of  functions  into  a  power 
series. 

For  when  the  error  term  in  Taylor's  and  in  Maclaurin's 
Theorem  approaches  zero  as  n  increases,  each  becomes  a 
convergent  infinite  series,  called  Taylor's  series  and  Mac- 
laurin's series  for  the  given  function,  about  the  given  point 
x  =  a. 

Some  functions  may  be  expanded  by  division,  some  by 
the  binomial  theorem,  others  by  the  logarithmic  or  the 
exponential  series.  All  of  these  series  are  but  particular 
cases  of  Taylor's  Theorem. 

211.  Another  Method  of  Deriving  Taylor's  and  Mac- 
laurin's Series.  —  1.  Maclaurin's  Series.  —  If  a  function 
of  a  single  variable  is  expanded  or  developed  into  a  series 
of  terms  arranged  according  to  the  ascending  integral  powers 
of  that  variable,  and  the  constant  coefficients  found,  the 
development  will  be  the  form  of  Maclaurin's  Theorem  with- 
out the  remainder.  Thus,  let  /  (x)  and  its  successive  deriv- 
atives be  continuous  in  the  neighborhood  of  x  =  0V  say 
from  x  =  —a  to  x  =  a,  and  assume  that  for  values  of  x 
within  that  interval, 

f(x)  =  A  +  Bx  +  Cx2  +  Dxz  +  Ex"  +  •  •  •  (1) 

If  equation    (1)    is   identically  true,   then   the  equation 
resulting  from  differentiating  both  its  members,  viz., 
f'(x)  =  B  +  2Cx  +  3Dx2  +  4:Ex:i  + 

also  is  identically  true  for  values  of  x  in  some  interval  that 
includes  zero.     For  similar  values  of  x,  the  following  equa- 


ANOTHER  METHOD  OF  DERIVING  SERIES        409 

tions  resulting  from  successive  differentiations  are  identically 
true: 

/"(*)  =  2C  +  2'SDx  +  S'±Ex2  + 

f'"(x)  =  2'3D  +  2>S'4;Ex+  •  •  ♦  . 

Py(x)  =  2-3-4E  + 

Putting  x  =  0  in  each  of  these  equations  gives : 
4=/(0),B=/mC=qp,D=qf^=«ete. 
Hence,  on  substituting  these  values  in  equation  (1), 
/(z)=/(0)+/'(0)*+^^  +  ^g!£!+  •  •• 

+/^r+--^  '  (2) 

which  is  Maclaurin's  series  as  in  (3),  Art.  209,  without  the 
remainder. 

If  f{x)  is  not  continuous,  no  development  according  to 
powers  of  x  is  possible.     Thus  if  /  (x)  =  log  x,  f  (0)  =  —  oo . 

A  power  series  represents  a  continuous  function,  hence  no 
power  series  in  x  can  be  expected  to  develop  log  x. 

It  is  evident  that,  whenever  the  function  or  any  one  of  its 
derivatives  is  discontinuous  for  x  =  0,  the  function  cannot 
be  developed  in  a  Maclaurin's  series. 

2.  Taylor's  Series.  —  Let  /  (x)  and  its  successive  deriva- 
tives be  continuous  in  the  neighborhood  of  x  —  a,  say  from 
x  =  a  —  h  to  x  =  a  -\-  h,  and  assume  that  for  values  of  x 
near  x  =  a, 

f(x)  =A+B(x-a)+C(x-a)2  +  D(x-ay 

+  E(x-ay+...,  (3) 

is  an  identically  true  equation. 

Then  the  following  equations  resulting  from  successive 
differentiations  are  identically  true  for  values  of  x  near  x  =  a. 


410  INTEGRAL  CALCULUS 

f  (x)  =  B  +  2C (x-  a)  +3D(x  -  a)2  +4E (x -  a)*  +  •  ■  ■  . 

f"(x)  =  2C  +  2-SD(x-a)+3'4:E{x-a)2+  •  ■  ■  . 
/"'(a;)  =+2-3D  +  2.3-4E(a;-a)+  •  •  •  . 
pv(x)  =  +2-3-4E  + 

Putting  x  =  a  in  each  of  these  equations  gives 
Hence,  on  substituting  these  values  in  equation  (3), 

/m  -/(«)  +/-rf  (*  - a)  +*^r  (*  -  °)s  +  •  •  • 

+  £M(x-a)"+  ■  ■  ■  ,  (4) 

which  is  Taylor's  series  as  in  (1),  Art.  207,  without  the 
remainder. 

If  in  (4)  x  is  put  for  a  and  (x  +  h)  for  x,  it  becomes 

f(x  +  h)=f(x)+f'(x)^+f"(x)^+  ■  ■  ■ 

which  is  Taylor's  series  as  in  (2),  Art.  208,  without  the 

remainder. 

Here  the  development  is  not  according  to  powers  of  x,  but 

of  some  value  (x  —  a)  or  h  near  to  the  value  x  =  a.      Hence, 

when  the  values  of  the  function  and  all  its  derivatives  are 

known  or  can  be  found  for  some  one  value  of  x,  say  a,  then 

the  value  of  the  function  for  x  =  a  +  h  can  be  found  from 

the    development.     Thus,    when   /  (x)  =  logo:;    /  (1)  =  0, 

f{1)  =  1,    /"(I)  =  -1,  /"'(l)  =  2!,  .  .  .  /(«)(!)  .  (-l)"+i 

(n  —  1)!;  and  the  series  will  be 

h2      h3      hA 
log(l  +  fc)=fc-J  +  J-j+ 

which  agrees  with  (1),  Ex.  1,  Art.  203. 


EXPANSION   BY  THEOREMS  411 

212.   Expansion  by  Maclaurin's  and  Taylor's  Theorems. 

—  If,  on  applying  to  a  given  function  any  one  of  +hese 
formulas,  the  last  term  becomes  0,  or  approaches  0  as  a  limit 
when  n  becomes  infinite,  the  formula  develops  this  function; 
if  not,  the  formula  fails  for  this  function.  That  is,  if  Rn  (x) 
=  0  when  n  =  x ,  Maclaurin's  or  Taylor's  series  is  the  devel- 
opment of /(x)  or/(x  +  h),  respectively. 

If/'1  (x)  increases  (or  decreases)  from  /n(0)  to/n  (x),  and  the 
sum  of  the  first  n  terms  in  Maclaurin's  series  is  taken  as  the 

value  of  /  (x) ,  the  error,  being  fn  (6x)  — . ,  lies  between 

/•(0)Jj     and    />(*)fj- 

If  fn  (x)  increases  (or  decreases)  from  fn  (x)  to  fn  (x  +  h),  and 
the  sum  of  the  first  n  terms  in  Taylor's  series  is  taken  as  the 

value  of  /(x  +  h),  the  error,  being/"  (x  -\-  6h)  —.,  lies  between 
f(x)h^    and    f(x  +  h)^- 


SY*7l  /v»         /*•         /y*  jy*  /y  /y» 

213.  Since  —  =  -.-.-     .  . -.-, 

n\      12  3  n  —  2  n  —  1  n 

xn 

— i  =  0  w/ien  x  is  finite  and  n  =  oo , 

for  last  factor  approaches  zero. 

Hence,  #„=*  (x)  =  0  when  fn  (dx)  is  finite,  or  when 
/"  (x  +  Bh)  is  finite,  in  Maclaurin's  or  Taylor's  Theorems, 
respectively. 

214.  If  /  (x)  =  /  (  —  x),  the  expansion  of/  (x)  will  contain 
only  even  powers  of  x;  while  if  /  (x)  =  —f(—x)  the  expan- 
sion of/  (x)  will  involve  only  odd  powers  of  x.  For  examples, 
see  the  expansions  of  sin  x  and  of  cos  x  following. 


412  INTEGRAL  CALCULUS 

215.   Examples.  —  1.  Sin  x.  —  Expansion  by  Maclaurin's 
Theorem : 

f(x)  =  smx,  /(0)  =sinO  =  0, 

/'(*)  =  cos  z,  /'(0)  =  1, 

/"(*)  =  -sin*,  f(0)=0, 

f'"(x)  =  -cosz,  /,,,(0)  =  -1, 

f"(x)  =  sinz,  f*  (0)  =0, 


fn(x)  =  sm(x+7fj,  /«(0)=sing), 

f»(6x)=8m(ox  +  ™y 

Since  sin  f  -^  J  is  0  or  ±1  according  as  n  is  even  or  odd,  the 

coefficients  of  the  even  powers  of  x  will  be  zero,  and  only  odd 
powers  of  x  will  occur,  the  terms  being  alternately  positive 
and  negative.     Thus, 

x3   ,  x5      x1   ,  ,  xn  .    I n     ,  nir\ 

smx  =  x-  -  +  --_+  •  •  •  +_sin(to  +  -J. 

Here,  Rn  (x)  =  —  sin  (dx  +  ~  ) , 

xn 
not  numerically  greater  than  —.,  which  has  zero  for  limit; 

.'.     R  (x)  =  0.     Hence  the  series 


n=oo 


/y»3  /y»5  /y»7  /v«9 

S1M  =  I_3!  +  5!"7!  +  9!"  "  "  '  (1) 

is  absolutely  convergent  for  every  finite  value  of  x. 

The  series  converges  rapidly  and  may  be  used  for  comput- 
ing the  natural  sine  of  any  "angle  expressed  in  radians. 

Thus,  for  the  sin  (5°43'46".5  =  TV  radian), 

sin  (0.1  radian)  =  0.1  -  ^^-  +  %r-5 =  0.09983 

o!  5! 


EXAMPLES  413 

For  the  sin(l°  =  ^  radian  =  0.017453  .  .  .  Y 

sinl°  =  sin(iio)  =  iio-(iio)33L! 

+(ifo)05i-"-=a017452---; 

.*.     sin  1°=  arc  1°,  to  five  places  of  decimals.     (See  Art.  40.) 

'  2.   Cos  x.  —  Expansion    may    be    made  by    Maclaurin's 

Theorem  as  is  done  for  the  sin  x,  or  the  differentiation  of  the 

sine  series  term  by  term  gives  the  series 

x2   .  x*      x6   ,   x8 

cos*=1-2!+4!~6!  +  8!~  "  "  '  (2) 

which  is  absolutely  convergent  for  every  finite  value  of  x. 

For  the  cos  (5°43'46".5  =  TV  radian), 

CO  l)2      (0  1Y 
cos  (0.1  radian)  =  1-^^-  +  ^^--  ■  •  -  =  0.995007 

For  the  cos(l°  =  ^radian  =  0.017453  .  .  . 

cosl°  =  cos^)  =  1  -  (I^)^  +  (i|g)4^ 

=  0.999847  .  .  .   ; 
/.    cos  1°  =  1  -  0.00015  .  .  . 

3.    Sin  (x  +  h).  —  Expansion  by  Taylor's  Theorem.    Here 
/  (x  -f  h)  =  sin  (x  +  h) ; 

f(x)  =  sinz,    f(x)  =  cosx,    f"(x)  =  —  sin.r,  etc. 

Hence, 

h2  h3  h4 

sin(x-\-h)  =  sinz+ftcosx  —  — f  sinx—  ^cosx+  — f  sinx+  •  •  • 

.       A       h2      h±      fc«  \ 

=  ™a5(1-2!  +  4!~6!+  •  "  ') 

+  COBa:(A"3!  +  5!~7!+  '  '  ')  (1) 

=  sin  x  cos  h  +  cos  x  sin  h,    (h  for  x  in  Exs.  1  and  2) 
the  well-known  relation  true  for  all  values  of  x  and  h. 


414  INTEGRAL  CALCULUS 


h3      h5      h7 
When  x  =  0  in  (1),   sin  h  =  h  —  oy  +  Fj-  7J  + 


as  in  Ex.  1  for  x.     The  series  (1)  is  rapidly  convergent  for 
small  values  of  h.     r. 
=  34'22".65;  then, 


7T  1 

small  values  of  h.     Thus,  let  x  =  -  and  h  =  -r^  of  a  radian 

6  100 


.  /f,    iv     ■  *7i      (0.0: 

SmU  +  100;  =  Sm6V1 2! 


(0.01)2      (0.01)4 


-(0.01   -»!  +  «!-...   ); 


4! 

6 

sin  30°  34'22".65  =  0.5  (0.99995  .  .  .  )+ 0.86603  .  .  . 
(0.00999  ....)=  0.50863  .  .  .  =  0.5 .+  0.00863  .  .  . 
4.    Cos  (x  +  h).  —  Expansion  may  be  made  in  the  same 
way  as  for  sin  (x  -f-  h),  or  the  differentiation  of  the  series  (1), 
h  being  constant,  gives  * 

cos  (x  +  A)  =  cosa;  ^1  -  2j  +  jj  -  gj  +  •  •  •  1 


sin  a; 


A       ft3      ft5      ft7   ,  \  ,Q. 


=  cos  x  cos  ft  —  sin  x  sin  ft,  the  well-known  relation. 

ft2      ft4      ft6 
When    x  =  0   in    (2),    cos/i=1_2l+4|-g]+  '  "  *  > 

as  in  Ex.  2  for  a\ 

When  x  =  !  and  ft  =  j~of  a  radian  =  34'  22,,.65,  in  (2); 

then     cos60°34'22".65  =  0.5  (0.99995  .'..)-  0.86603  .  .  . 

(0.00999  .  .  .  )  =0.49132  .  .  .  =0.5-0.00868 

5.   ax  and  ex. — Expansion  by  Maclaurin's  Theorem.    Here 

*          /(*)-*■,  /(0)  =  a°=l, 

f(x)  =  a*  log  a,  /'(0)  =  loga, 

/"(x)  =  a*(loga)2,  /"(0)  =  (log  a)2, 

f'"(x)=a>(\ogay,  f"'(0)  =  (log  a)3, 


fn  (x)  =  a*  (log  a) »,  /» (0)  =  (log  a) », 

fn(6x)  =  afe(loga)\ 


EXAMPLES  415 

a  log  a   ,   (a;  log  a)2      (a:  log  a)3 
•'     a=1+_T"+       2!       +       3!       +  '" 

n\ 
Since  fn  (x)  =  ax  (log  a)n,  when  a  is  positive,  f(x)  and  all  its 
successive  derivatives  are  continuous  for  all  values  of  x. 


When  a;  is  finite,  ate  is  finite.     By  Art.  213,  v     "°    }    =  0, 


(xloga)n  ±_ 
~n\ 

when  n  =  oc  and  x  log  a  is  finite.     Hence  Rn  (x)  =  0  when 
n  =  ob'  and  z  is  finite.     Therefore,  the  exponential  series 
xbga      (xlogo)^  (slog  a)""1 

"*■      1      ^       2!       "r"    "r    (n-1)!    -1  l  ; 

is  the  development  of  ax  when  a  is  positive  and  z  is  finite, 
being  then  absolutely  convergent. 

Value  of  ex.  —  Putting  a  =  e  in  (1),  gives  (since  log  e  =  1) 

5^1)1 

Fcrfwe  0/  e.  —  Putting  x  =  1  in  (2),  gives 

*=1  +  1  +  2l  +  l!+i+-'-+(^!+--- 
=  2.718281  ....     (See  Art.  34.) 

6.   LogJ(l  +  #). —  Expansion  by  Maclaurin's  Theorem. 
Here 

/  (x)  =  loga  (1+  x),  f  (0)  =  loga  1  =  0, 

..,  .         m 


'-i+f  +  21  +  il  +  li-  Nlf^lTT-  w 


1  w_l+*' 

./    \vj  -  in, 

/"W  =  -(TT^' 

/"(0)  -  -m, 

>"'(*>  =  (1+V 

/'"(<))  =  2  m, 

416  INTEGRAL  CALCULUS 

(—l)n-l  (n  _   l)\m 

fn(x)  -  (i+g)»        '      /" (°)  =  ^"1)n_1  ^  ~  W™* 

;  {m  (i  +  dx)n 

:.     loga  (1  +  x) 

(        x2   ,  z3      a;4  .  .   (x-6x\n  f(-l)n-l\ 

When  x  <  —  1,  loga  (1  +  x)  has  no  real  value. 
When  x  =  —  1,  the  odd  derivatives  are  discontinuous. 
When  x  >  —l,f(x)  and  all  its  successive  derivatives  are 
continuous. 

Cauchy's  form  of  remainder, 

fl,(»)=/-(te)8^_g|    > 

gives  B.(X)  =  ^__J   --^^-m, 

in  which  the  second  factor  is  finite,  and  the  first  factor  =  0, 
when  n  =  oo  and  x  >  —  1  and  <  1  or  x  =  1. 
Hence  the  logarithmic  series 

l0ga  (1  +  X) 

(         X*X*       X*.  ,    (-l)*-g3»-l    ,  \ 

is  the  expansion  of  log0  (1  +  x)  when  #  >  —  1  and  <  1  or 
x=l. 

Putting  —x  for  x  in  (1),  gives 

loga(l-z)  =  m(-z-  |-|-|4+  •  •  •  ).         (2) 

Subtracting  (2)  from  (1),  gives 

.       l+x     0     /    .  a:3      a:5  .  x7  .         \  /Compare  (3),  \     ,Q* 
loga  j-j  =  2m(x+  -+-+-+...  J.  ^  ^  203J    (3) 

T     i  !  +1  1  +Z         Z+  1  fAS 

Let         a:  =  27+T;  thenr^  =  — •         (4) 


EXAMPLES  417 

Substituting  in  (3)  the  values  in  (4),  gives 

log/-±i  =  2m  (^  +  3(2/+1),  +  •••);      (5) 
••     loga(^+l)  =  logeg+2m(^I  +  3(221+1)3+...).    (6) 

When  2>0,  0<£<1;  hence  the  series  in  (6)  is  conver- 
gent for  all  positive  values  of  z. 

When  1  is  put  for  z,  loga  2  is  found ;  then  2  for  z,  and  loga  3 
is  found;  thus  loga  (z  +  1)  can  be  readily  computed  when 
loga  z  is  known.     (See  Ex.  1,  Art.  203.) 

Whenra  =  1,  a  =  e,  the  Napierian  base;  thus  (5)  becomes 


log"-Ti  =  2(2iTl  +  sW+W  +  "  '  •  )• 

2  +  1 

Dividing  (5)  by  (7)  and  denoting  — —  by  N,  gives 


(7) 


logaAVlogN  =  m;  that  is,  logaN  =  mlogiV.         (8) 

Valve  of  m.  —  Putting  N  =  a  in  (8),  gives 

1  =  mloga;  that  is,  m  =  1/loga,  (9) 

or  N  =  e,  gives,  logae  =  m\ 

:.     1/loga  =  loga  c,    or     1  =  log  a  •  loga  e.         (10) 

Valve  of  M.  —  Denoting  the  modulus  of  the  common 
system  whose  base  is  10  by  M;  from  (9)  and  (10), 

M  =  io^io  =  logloe 

0.434294  ....     (Compare  Art.  38.) 


2.302585 


Note.  —  When  a  =  10  and  m  =  M,  or  when  a  =  e  and 
m  =  1,  all  the  series  in  this  example  become  common  loga- 
rithmic series,  or  Napierian  logarithmic  series,  respectively. 


418  INTEGRAL  CALCULUS 


0-i<i)- 


(by  combining  terms\ 
of  5  and  6.         / 


EXERCISE   XLm. 

,     .  *3  ,   2x*  ,    17x7, 

1.  tanx  =  x+3+1J  +  w  + 

0  1    .  x2  ,   5x4  ,  161  X*   , 

2.  secz  =  l  +  2+^4+i-720]+'--' 

X3    ,    X5         X1    ,     X9 

3.  tan-'*  =  z-3  +  j-y+g . 

.      .     ,  ,   1    x3  ,   1-3    3?  ,   1-3-5    a?   , 

/v.2  /7*3  *v>4 

6.   e*  =  l+z  +  !|  +  3l  +  !]+.... 

o»2  /j*3  ^t^4 

6.  e~x  =  1  -  x  +  —  -  ^  +  -j  -  •  •  •  ,    by  replacing  x  by  -x  in  5. 

7.  sinh  x  = ^ —  =  x  +  q"|  +  51  +  * 

ex  +  e~x  x2      x4 

8.  cosh  x  = = =  ^  +  o7  +  41  +  '  '  '  >    ^y  combining  terms 

of  5  and  6,  or  by  differentiating  terms  of  7. 
A  ,  sinhx  x3  .   2X5      17  x7   . 

9-  tanha;  =  J35^  =  :c-3+T5--^l5-+---- 

10.  From  5  and  (1)  and  (2)  of  Examples  1  and  2,  making 

x  =  V—  1  •  x  =  ix, 
get  ei*  =  cos  x  +  i  sin  x,  (1) 

and  e-ix  =  cos  x  —  i  sin  x.  (2) 

Note.  —  Putting  t  for  x  in  (1)  gives  the  remarkable  relation,  el7r  =  —  1 ; 
while  putting  —  -w  for  x  in  (2)  gives  el7r  =  1,  hence  iir  is  an  imaginary 
value  of  log  1,  the  real  value  being  zero. 

11.  From  (1)  and  (2)  of  10,  by  subtraction  and  by  addition  get 

eix  _  e-ix  eix  _|_  e-ix 

sin  x  = — (3),  cosx  =  ^ (4) 

12.  Evaluate    f  e^  dx  =    f    (1  -  x2  +  ix4  -  ix6  +  •  •  •  ) dx. 

Get  result  when  end  value  is  1.  When  end  value  is  00  the  value  of 
the  integral  is  \  v^.*  This  integral  is  important  in  the  theory  of 
probability. 

*  Williamson's  Integral  Calculus,  Ex.  4,  Art.  116,  also  Gibson's 
Calculus,  Ex.  3,  Art.  136. 


THE   BINOMIAL  THEOREM  419 

216.  The  Binomial  Theorem.  —  I.  The  binomial  theorem 
is  seen  to  be  a  special  case  of  Taylor's  Theorem  by  expanding 
(x  +  h)m  in  a  power  series  in  h  by  Taylor's  formula.     Thus, 

f(x  +  h)  =  (x  +  h)™;  .'.     f(x)=xm, 

f  (x)  =  mxm~1,  f  "  (x)  —  m  (m  —  1)  xm~2, 

/'"  (x)  =  m  (m  -  i)  (m  -  2)  xm~3, 


fn~1(x)=  m  (m  —  1)  .  .  .   (m  —  n  +  2)  xm~n+1. 
Substituting  these  values  in  Taylor's  series  (5),  of  Art.  211, 

(x  +  h)m  =  xm  +  mxm~l  h  +  m(ra~  1)a;m-2^2 

+  m  (m  -  1)  (m  -  2)  ^m_3/i3  +  .  .  . 
o . 
.  ,  .       m(m-l)     .  .(m-«  +  2)x„_„+1/t„_1+  .  .  .      ^ 
(n  —  1)! 

is  the  resulting  Binomial  Theorem. 

Here    fn  (x)  =  m  (m  —  1)  .  .  .  (m  —  n  +  1)  zm-n. 

Hence,  /  (x)  and  all  its  successive  derivatives  are  continu- 
ous for  all  values  of  x. 

hn(\  —  6)n~l 
Cauchy's  form  of  remainder,  Rn  (x)  =fn  (x+dh)     .    _n>     , 

gives 


m  (m  -  1)  .  .  .(m-n  +  1)    (h  -  dh\n    (x  +  dh)m 
KnW-  (n-  1)!  \x  +  0h)   '      1-0 

When  |  X  \  >  h  and  n  =  oo ,  the  product  of  the  first  and 
second  factors  =  0,  and  the  last  factor  is  finite;  hence, 
R  (x)  =  0. 

71=00 

Hence,  the  binomial  theorem  holds  true  when  the  first 
term  of  the  binomial  is  greater  absolutely  than  the  second. 

When  m  is  a  positive  integer,  the  series  (1)  stops  with  the 
(m  +  l)th  term,  since/71  (x)  =  0  when  n  >  m,  and  is  therefore 
a  finite  series  of  m  +  1  terms, 


420  INTEGRAL  CALCULUS 

If  |  h  |  >  x,  the  expansion  may  be  a  power  series  in  x ;  thus, 

/7    ,     n         7      ,       71       t  m(m—  1)  hm~2   .   . 

(h  +  x)m  =hm  +  nth™'1  x  -\ x?   x  +  *  '  ' 

m(m-l).  .  .  (w  -  n  +  2)  fe"-"+1       ,  ,  . 

...+  ^— ^  x      +  ...      (2) 

is  a  true  expansion  when  \h\>  x. 

II.  (1  -f  x)m  may  be  expanded  in  a  power  series  in  £  by 
Maclaurin's  Theorem,  giving 

/-,    ,     n       -,    ,  i  m(m—l)   ,  .  m(m—  l)(m  —  2)    ,  . 

(1  +  a:)»  =  1  +  ?nx  +        2|    ,  V+  — * ^ -x*  +  •  •  ' 

.  .       ■  ^(m-  1)  .  .  .  (m-n  +  2) 

.-.+  (n_i)!  x      +  -W 

As  in  case  I,  when  m  is  a  positive  integer,  the  series  (3) 
stops  with  the  (rn  +  l)th  term  and  is  therefore  a  finite  series 
of  m  +  1  terms.  If  m  is  negative  or  fractional,  the  series  is 
infinite.  The  ratio  test  shows  that  the  infinite  series  con- 
verges absolutely  when  |  x  \  <  1  and  diverges  when  \x\>  1 ; 
therefore  Rn  (x)  needs  examination  only  f or  |  x  |  =  1. 

Here    fn  (x)  =  m  (m  —  1)  .  .  .  (m  —  n  +  1)  (1  +  x)m~n, 
fn  {Bx)  =  m  (m  -  1)  .  .  .  (m  -  n  -f  1)  (1  +  Bx)m-n. 

xn  n    _  0)n-l 

Cauchy's  form  of  remainder,    Rn  (x)  =fn  (dx)  — t -y. — , 

gives 

For  values  of  x  between  0  and  ±1,  the  last  factor  is  finite 
for  every  n ;  the  second  factor  is  always  positive  and  cannot 
exceed  unity;  the  first  factor  approaches  zero  as  a  limit  as 
n  increases  without  limit,  since  it  is  the  expression  for  the 
nth  term  of  the  convergent  series 

l  +  (m-l),  +  (w-1)2jm-2)^+---. 


APPROXIMATION   FORMULAS  421 

Hence,  R  (x)  =  0,    and   the  infinite   series   converges  to 

n=oo 

(1  +  x)m  for  every  value  of  m,  when  |  x  |  <  1. 

For  x  =  dbl,  the  following  results  may  be  found 
proved  in  Chrystal's  Algebra.  These  cases  are  not  so 
important. 

When  x  =  +1,  the  series  converges  absolutely  if  m  >  0, 
but  conditionally  if  0  >  m  >  —  1,  oscillates  if  m  =  —  1, 
and  diverges  if  ?n  <  —  1 . 

When  x  =  —  1,  the  series  converges  absolutely  if  m  >  0, 
and  diverges  if  m  <  0. 

If  a  >  b,  (a  +  b)m  can  be  written  am  (1  +  b/a)m  and 
expanded  by  Maclaurin's  formula,  since  b/a  <  1  may  take 
the  place  of  x  in  (1  +  x)m.     Hence,  in  this  case, 

(a  +  b)m  =am  +  mam~l  b  +  -  ^^  1}  am~2b2  +  ■  •  •  ,    (4) 

which  agrees  with  (1)  and  is  the  Binomial  Theorem,  proved 
true  for  a  >  b  whether  m  is  positive  or  negative,  whole  or 
fractional. 

If  a  <  b,  interchange  them  in  (4)  and  the  result  will  agree 
with  (2)  and  be  a  true  expansion  of  (b  +  a)m. 

217.  Approximation  Formulas.  —  Often  a  function  may 
be  replaced  by  another  having  approximately  the  same 
numerical  value  but  a  form  better  adapted  for  computation. 
In  such  cases  the  given  function  may  be  expanded  in  a  series 
and  a  certain  number  of  terms,  beginning  with  the  first, 
taken  as  an  approximate  value  of  the  function;  the  number 
of  the  terms  taken  being  according  to  the  precision  desired 
for  the  result. 

The  binomial  theorem  furnishes  one  of  the  most  useful  of 
the  approximation  formulas.  Thus,  if  m  denotes  a  small 
fraction,  expanding  (1  ±  m)n  gives 

(1  dbf»)»  =  1  ±  nm  +  n  {n~  1}  m2  ±  •  •  •  , 


422  INTEGRAL  CALCULUS 

where,  since  m  is  small,  neglecting  powers  higher  than  the 
first,  the  approximate  relation, 

(l±m)n=  1  ±nm  (1) 

results.     For  the  special  case  n  =  J, 

Vl±m  =  1  ±  |  m.  (2) 

For  6  small  in  comparison  with  a,  the  general  form  is 

V?±&  =  a(l±^)-  (3) 

For  examples:  Vl  +x  =  l  +  Jx—   •  •  •   ; 

=  1  —  x  +  •  •  •  ;       ,  =  1  —  -x  +  •  •  •  . 


1+z  r    Vl+x  2 

For  extraction  of  roots  in  general 


i 
(an±  b)n=  ail  ±—  J    =a(l±x)n,  (4) 


7  l 

where  x  =  — .     Expanding  (1  ±  x)n  gives 

(l.^T-ljA     (n-D^x(n-l)(2n-l)^ 

(i±xj  -i±n«       n2    2!±  n3  3!  •    (s; 

Example.  —  v'lOOO  =  ^1024  -  24  =  4  (1  -  Tyi 
Substituting  Tf  ^  for  x  in  the  series 

(  )0  5      5  10      5  10  15 

gives  to  six  figures  0.995268;  hence, 

v/T000  =  4  X  0.995268  =  3.981072. 

Since  ex  =  1  +  x  +  — .  +  —.+  •  •  •  ,  when  x  is  small 

e*  =  1  +  x  (6) 

is  the  approximate  relation. 


APPROXIMATION   FORMULAS 


423 


From  the  series  for  sin  x,  cos  x,  and  log  (1  +  x), 
sin  x  =  x  (1  —  \x2)} 
cos  x  =  1  —  \  x2, 
log(l  +z)  =  x  -  \x2, 
are  the  approximate  relations  when  x  is  small. 
When  x  is  small  compared  with  a, 

sin  (a  ±  z)  =  sin  a  d=  £  cos  a, 

log(a  +  x)  =1°ga+^-2^' 


1 


1       #       a;2 

-  -F  -^  H — ^' 

a      a2      a3 


(7) 
(8) 
(9) 


(10) 
(11) 

(12) 


a  ±x 

are  approximate  relations  when  succeeding  terms  of  the 
respective  series  are  neglected.  -^ 

In  all  these  cases  the  error  made  in  taking  the  approxima- 
tion for  the  value  of  the  function  may  be 
closely  estimated  from  the  value  of  Rn  (x), 
the  error  term,  for  the  particular  series  em- 
ployed. 

Example.  —  In  considering  the  length  of  a 
circular  arc  and  its  corresponding  chord  in 
railway  surveying,  use  may  be  made  of  the 
approximate  relation  (7).  Thus,  letting  s 
denote  length  of  the  arc,  r  the  radius,  c  the 

chord,   a  the  angle  in   radians;    s  =  ra   and 

When  a  is  small, 

-»'*i-»i|>-gS)l 


2  r  sin 


—  ra  —  3*5  ra6 
or  for  a  in  degrees,  s  —  c  = 


s  —  c 


raA 


4514180' 


s  4  re? 
Since 


the  error  of  the  approximation  cannot  exceed 


ra° 
1920 


424  INTEGRAL  CALCULUS 


EXERCISE   XLIV. 

1.   Expand  (x  +  y)m.  2.   Expand  (x  +  yf. 

3.   Expand  ex+h.  4.   Expand  log  sin  (x  +  h). 

6.  Expand  sin"1  (x  +  h).  6.    Expand  eainx. 

7.  Given  /  (x)  =  x3  -  4  x  +  7,  find  /  (x  +  3)  and  /  (x  -  2)  by 
Taylor's  series.  Then  find  /  (x  +  3)  and  /  (x  —  2)  by  usual  algebraic 
method  and  thus  verify  results. 

8.  Using  the  approximation  formula  (12)  compute  the  reciprocal 
of  101;  and  of  99.     Compare  results  with  those  obtained  by  division. 

9.  Find  the  length  of  the  chord  of  an  arc  of  radius  5729.65  feet 
subtending  an  angle  of  1°:  (a)  by  trigonometric  methods;  (6)  by  the 
approximation  formula  (7).  Find  results  when  the  radius  is  5729.58 
feet.     Compare  results  and  find  error  of  approximation. 

10.  Find  the  length  along  the  slope  of  a  road  that  rises  5  ft.  in  a 
horizontal  distance  of  100  ft.  by  the  approximation  formula  (3).  Deter- 
mine to  how  many  places  of  decimals  is  the  result  correct. 

218.  Application  of  Taylor's  Theorem  to  Maxima  and 
Minima.  —  This  Article  is  supplementary  to  Art.  83,  being 
an  additional  proof  of  the  rule  given  in  that  Article  for  the 
determination  of  whether  a  critical  value  x  =  a,  a  root  of 
f'(x)  =  0,  makes  f(x)  a  maximum,  a  minimum,  or  neither. 

Let  /  (x)  be  a  function  of  x  such  that  /  (a  +  h)  and 
f  (a  —  h)  can  be  expanded  in  Taylor's  series,  and  let  /  (a)  be 
the  value  to  be  tested. 

Developing  f  (a  —  h)  and  /  (a  +  h)  by  formula  (2),  Art. 
208: 

f(a-h)=f(a)-hf(a)+^f'>(a)-p'"(a)+  •  .  . 

f(a  +  h)  -  /(a)  +  hf  (a)  +  g/(a)  +  J/'". (a)  +  •  •  • 

+  y$lf(a  +  <W)l     (2) 

in  which  0i  and  82  are  between  0  and  1  in  value. 


INDETERMINATE  FORMS  425 

When  the  first  n  —  1  derivatives  of/  (x)  are  zero  for  x  =  a 
and  the  nth  derivative  is  not  zero  for  x  =  a,  then, 

f(a  -  h)  -/(a)  =  t_^->  (a  -  6,h),  (3) 

/(a  +  *)-/(«)-j^(a  +  «k).  (4) 

Since  /  (z)  and  its  successive  derivatives  are  assumed  to 
be  continuous  at  and  near  x  =  a,  the  signs  of  /"  (a  —  0i/i) 
and  /"  (a  -+-  02ft),  for  very  small  values  of  ft,  are  the  same  as 
the  sign  of  fn  (a) . 

It  is  manifest  that  if  n  is  an  even  integer,  /  (a)  will  be  a 
maximum  or  a  minimum  according  as  fn(a)  is  negative  or 
positive;  and  if  n  is  odd,  f  (a)  will  be  neither  a  maximum  nor 
a  minimum  whether  the  sign  of  /"  (a)  is  negative  or  positive. 

These  conclusions  are  manifest  because  when  n  is  even  and 
fn(a)  is  negative,  the  left  members  of  (3)  and  (4)  are  both 
negative,  and  hence  f  (a)  >  f  (a  —  h),  f  (a)  >  f  (a  +  h); 
that  is,  /  (a)  is  a  maximum. 

When  n  is  even  and  fn  (a)  is  positive,  the  left  members  are 
both  positive,  and  hence/  (a)  <  /  (a  —  h),f  (a)  <  /  (a  +  h) ; 
that  is,  /  (a)  is  a  minimum. 

When  n  is  odd,  whether  fn  (a)  is  negative  or  positive,  the 
left  members  have  different  signs,  and  hence/  (a)  >/  (a  —  /*), 
/  (a)  >/  (a  +  ft) ;  that  is,  /  (a)  is  neither  a  maximum  nor  a 
minimum  regardless  of  the  sign  of  fn  (a). 

219.  Indeterminate  Forms.  —  It  was  noted  at  the  end 
of  Art.  20  that  the  derivative  of  f(x), 

may  be  finite,  zero,  or  non-existent,  but  not  0/0. 

The  symbol  0/0  is  called  an  indeterminate  form,  and 
when  /  (x)  takes  that  form  for  some  value  of  x,  say  a,  then 
/  (x)  is  really  undefined  for  x  =  a,  although  it  may  be  defined 
for  any  other  value  of  x.     It  is  possible,  however,  that  /  (x) 


426  INTEGRAL  CALCULUS 

may  have  a  definite  limit  A  when  x  converges  to  a;   it  is 

customary  then  to  call  /  (a)  =  0/0  an  indeterminate  form, 

and  to  define  A  as  the  value  of  /  (x)  when  x  =  a,  calling  it 

the  true  value  of  /  (x)  for  x  =  a. 

The  advantage  of  having  this  "true  value"  assigned  by 

definition  is  that/(V),  being  in  general  continuous,  thereby 

becomes  continuous  up  to  and  including  the  value  a. 

x2  —  4 
Take,  for  example,  the  function  y  =  — .     For  every 

X      z 

value  of  x  other  than  x  =  2,  the  function  has  a  definite  value, 

4  —  40 

but  for  x  =  2  it  becomes  - -  =  ^ .     Since  the  function  has 

£  Ll  U 

no  definite  value  when  x  =  2,  the  limit  which  the  function 
approaches  as  x  converges  to  the  value  2  is  assigned  as  the 
value  of  the  function  when  x  =  2.     If 

h=o    I  -\-  h  —  A  h±Q 

x2  —  4 
.\     lim =  4. 

x±2    x  -  2 

Thus  the  true  or  limiting  value  of  this  function  which  takes 
the  indeterminate  form  0/0  is  4. 
For  values  of  x  other  than  2, 

.-.      X—±        =  lim  (x  +  2)  =  4. 

X  —  1  JJ=2         x=2 

On  the  graph  of  y  =  x  +  2,  the  ordinates  of  points  for 
values  of  x  other  than  2  represent  the  values  of  the  function, 
but  for  x  =  2,  the  function  having  no  definite  value  may  be 
represented  by  any  ordinate  lying  along  the  line  x  =  2.  Of 
the  values  that  may  be  assigned  to  the  function  for  x  =  2, 
there  is  one  value  represented  by  MP  =  4,  which  is  the 
limit  of  the  values  represented  by  the  ordinates  of  points  on 


INDETERMINATE   FORMS 


427 


y  =  x  +  2  as  x  approaches  2;    and  it  is  desirable  to  select 
this  value  of  y  as  the  value  of  the  function  when  x  =  2. 

By  this  selection  the  function  is  defined  for  x  =  2  and  thus 
becomes  continuous  through  that  value  of  the  variable  x. 


— x 


In  general,  lim  /  (x)  defines  the  value  of  the  function  when 

x—a 

fix)  is  indeterminate  for  x  =  a.  The  expression  f (x)]a 
denotes  the  value  of  f(x)  when  x  =  a. 

The  value  of  a  function  of  x  for  x  =  a  usually  means  the 
result  obtained  by  substituting  a  for  x  in  the  function. 

When,  however,  the  substitution  results  in  any  one  of  the 
indeterminate  forms, 

0  0,     oo /oo,     0.x,     x  -  oo,     0°,     x 


1 


the  definition  must  be  enlarged;  thus,  the  value  of  a  function 
for  any  particular  value  of  its  variable  is  the  limit  which 
the  function  approaches  when  the  variable  approaches  this 
particular  value  as  its  limit . 

This  definition  need  be  used  only  when  the  ordinary 
method  of  getting  the  value  of  the  function  gives  rise  to  an 
indeterminate  form. 


428  INTEGRAL  CALCULUS 

220.  Evaluation  of  Indeterminate  Forms.  —  In  many 
cases  the  limits  desired  are  easily  found  by  simple  algebraic 
transformations  or  by  the  use  of  series.  When  the  function 
that  assumes  the  indeterminate  form  is  the  quotient  of  two 
polynomials,  or  can  be  put  in  that  form,  the  following  direc- 
tions may  be  of  service. 

fix) 

1.  If  the  function  is  of  the  form  t-t-(  and  becomes  0/0  for 

4>(x) 

x  =  0,  divide  both  numerator  and  denominator  by  the  lowest 
power  of  x  that  occurs  in  either.  If  the  fraction  becomes 
oo/oo  for  x  =  oo,  divide  both  terms  of  the  fraction  by  the 
highest  power  of  x  in  either. 

fix) 

2.  If  the  function  has  the  form     ,  ;  and  becomes  0/0  for 

4>(x) 

x  =  a,  divide  both  terms  by  the  highest  power  of  (x  —  a) 
common  to  both 


EXAMPLES 


x3  +  3  x2  —  5  x 


_5 
3x4-  2x3  +  6x|o  6' 

When  x  =  0,  this  fraction  takes  the  indeterminate  form 
0/0.  Hence  to  evaluate  it  for  x  =  0,  its  limit  when  x  =  0. 
must  be  found.     For  values  of  x  other  than  0, 

x3  +  3  x2  -  5  x  x2  +  3  x  -  5    . 


2. 


3x4-  2x3  +  6x      3x3-  2x2  +  b' 

x3  +  3x2-5x        ,.        x2  +  3x-5 
S  3x4  -  2x3  +  6x  ~  S2J  3x3  -  2x2  +  6 

5 
6 

x3  +  3  x2  -  5  x  1 
3x4-2x3  +  6x_L 

13       5 

x3  +  3  x2  —  5  x        x      x2      x3 . 
3x4-2x3  +  6x               2  ,6' 

X         X3 

EVALUATION  OF  INDETERMINATE   FORMS       429 
x3  +  3  x2  -  5  x 


£s,3 


i  +  i_5 

..      x^x2      x3      0      n 
■ lun  — 2~T  =3  =  °' 


3  -  -  +  -3 
x      x3 


3. 


vrr 


vr 


Jo 


By  rationalizing  numerator, 

Vl+x-  Vl-x  _  /Vl+x-  Vl-aA  /Vl+a  +  Vl-sA 

2 


vi  +  x  +  vi 


V    lim  r 

x=o  L 


Vl  +x-  VI 


J         x=0   - 


o  Vl  +  a;  + 


1-aJ 


4.   Vl  +  x  -  Vx 1    =  0. 
By  changing  form, 


vT+~x  -  Vx  =  (Vi+x  -  Vx) 

1  +  x  —  X 


(Vl+x  +  Vx) 
(Vl+x  +  Vx) 


Vl  +x  +  Vx' 
.     lim  (Vl+x  -  Vx)  =  lim      /T-r-   , — 7=    =  °- 

x  =  oo  *  =  x|_Vl  +  £  +  VxJ 

a;  —  sin  x~]        1 
"Jo  =  6 


x3 

By  expanding  sin  x  in  series, 

x  —  sin  x 
x3  x 

6      5!  +  7! 


lim' 

i=0  x3 


sinx 


?[x-{x-i+i. )] 

•  •  •  ,  if  *  *  0; 


430 


INTEGRAL  CALCULUS 


6. 


smx-x  cos  x 


1 


xc  JO 

By  expanding  sin  x  and  cos  x  in  series, 
sin  x  —  x  cos 


"~a?L\        3!  +  5! 

( 


x4    .or 
2!+4! 


z=0  L 


X  COS 


5]= 


)] 


lim 

x=0 


=  lim 

x=0 


-( 

X3\ 


30 


+ 


30  + 


)] 

B 


221.  Method  of  the  Calculus.  —  For  the  form  0/0,  to 
which  all  other  indeterminate  forms  may  be  reduced,  Tay- 
lor's Theorem  furnishes  a  general  method  of  evaluation. 

I.   When/(#)  and  0  (x)  are  continuous  functions  of  x  and 


reduces   to   the   form   0/0  for   x  =  a,   the   value  of 


lim 

2 


is  desired. 


(1) 


That  is,  if  £/ie  ratio  of  two  functions  of  x  takes  the  form  0/0 
when  x  =  a,  then  the  ratio  of  these  functions  when  x  =  a  is 
equal  to  the  ratio  of  their  derivatives  when  x  =  a. 

If  /  (x  +  h)  and  </>  (x  +  h)  can  be  expanded  by  Taylor's 
formula  in  the  neighborhood  of  x  =  a,  it  is  seen  that 

»-  (MWI  =  9f\ ■<  ^t  is,  \tm  =  pi .  (1) 

By  Taylor's  Theorem   [Art.   208,   formula   (2)],   putting 
a  for  z, 

/(o  +  fe)  =  fW+hf'ja  +  Oji)  =  /'(a  +  fl.ft) 
</>  (a  +  /*)       0  (a)  +  /i0'  (a  +  62h)      0'  (a  +  02/i) ' 
since /(a)  =  0,  4>(a)  =  0;  [See  also  (1„),  Art,  205] 


METHOD  OF  THE   CALCULUS  431 


+  W 


+  02h). 


fM.    (1) 


If /'(a)  and  0'  (a)  are  both  zero,  then  [(2a),  Art.  205] 


fc-o  |>  (a  +  h)  J       ft=o  U    (a  +  Oji)]       <t>"  (a 


In  this  way  it  is  seen  that  if,  for  z  =  a,  f  (x)  and  </>  (x)  and 
their  successive  derivatives,  including  their  nth  derivatives, 
are  zero,  while  fn+1  (a)  and  (f>n+1  (a)  are  not  both  zero,  then 


limr/j£)]=limr/!^Mi;  that  is,  \mi  jgm 


(2) 


fix) 

If  the  function     ,  \  takes  the  form  0/0  when  x  is  infinite, 
<f>{x) 

by  putting  x  =  -  the  problem  is  reduced  to  the  evaluation 
z 

of  the  limit  for  z  =  0,  and  hence  the  method  applies  to  this 

case  also. 

fix) 

II.   Form  go  /go  .  —  When  the  function  :^-4  takes  the  form 

oo  /qo,  it  can  be  reduced  to  the  form  0/0,  by  writing  it  in 

the  form  —7-^  /tt-t.     This  form  can  be  evaluated  as  before. 
4>(x)/  fix) 

Thus,  let  f(a)  =  go  and  </>  (a)  =  00,  a  being  finite  or  in- 

Now  2-^7  =  —p-r  /  77-r ,  which  is  in  the  form  0/0.     Ap- 

<f>(a)      0(a)/  /(a)' 

plying  formula  (1), 


432  INTEGRAL  CALCULUS 


r/Ml       [0W12  _  r/W T  £M  =  [7MT  R^l  • 
U(*)J.~    fw     U(A)J  '/'(«)    Uwl'L/'^J.' 


[/(«)? 


r/wi  .    jl  =  rrwi . 
Uc*oJ«   kwi    U'wJa 


If    ,,  {  is  indeterminate,  continue  according  to  formula  (2) 

until  two  derivatives  are  obtained  whose  ratio  is  deter- 
minate, which  ratio  is  the  limiting  value  sought  for  the 
function. 

III.  Other  Forms.  —  The  evaluation  of  the  other  indeter- 
minate forms  may  be  made  to  depend  upon  the  preceding. 

(a)  Form  0  •  oo  .  —  When  a  function  f(x)  •  <j>  (x)  takes  the 
form  0  •  oo  for  x  =  a,  it  may  be  reduced  to  the  form  0/0  or 
oo/oo  ;  thus, 

/».#«- *g    or     *g. 

(6)  Form  oo  —  oo .  —  By  some  transformation  and  simpli- 
fication, a  function  taking  the  form  oo  —  oo  may  be  reduced 
to  a  definite  value,  or  to  one  of  the  preceding  indeterminate 
forms. 

(c)  Forms  0°,  oo°,  l00.  —  These  forms  arise  from  a  func- 
tion of  the  form  [f(x)]^x).  This  function  may  be  reduced 
to  the  form  #.     Thus  let  y  =  [f  (x)]*(x),  whence 

\ogy  =  4>(x).\og[f(x)].  (3) 

Since  for  each  of  the  given  forms,  (3)  takes  the  form  O.oo, 
the  evaluation  is  effected  as  in  (a),  the  value  of  y  being 
found  from  log  y. 


METHOD  OF  THE   CALCULUS 


433 


EXAMPLES. 


x2-4 
x-2 


1- 


4. 


x2-41  =0. 
x-2j2      0' 


x-2 


2x1 
1  J2 


4.     (As  in  Art.  219.) 


x-sinx 


Jo      6 

sin  xl  _  0 .         .       x  —  sin  xl        1  —  cos  xl 
?       Jo  ~  0 '        "  x1       Jo  ~       3X2"      J0 

_  sin  xl   _  cos  xl    _  1 
6  x  Jo         6    Jo      6 

-  anl  . 

=  na"-1. 

-  a  ]a 

n  —  anl       0  xn  —  anl       nxn~ll  . 

=  n^         •'•      =  — \ —     =  na"-1. 

x  —  a  Ja      0  x  —  a  Ja  1     Ja 

.     ax  -  ¥1       .      a 
4.    =  logr- 

£       Jo  O 

a*  -  frxJ       °  •  a*  -  6*1       .  x      .      ,    .  1 

■ =  - ,        .'.      =  log  a  •  a1  -  log  6  •  b*  \ 

X       Jo       0  X       Jo  Jo 


xrt  — 
z 


log  a  —  log  b  =  log 


gx  _  e-x  _ 


x  — 

e  —  e~x  —  2  x 
x  —  sin  x 


1^£T  =  2. 

sin  x     Jo 

1  _  ?         .       ^  -  e~z  -  2  x]  _  e*  +  e~x  -  21 
Jo      0 '      "  x  —  sin  x     Jo         1  —  cos  x  J0 

^gx  -  e~xl  J&j±jrf\ 
sinx   Jo        cosx  J0 

UmLlog(l+«)l=?5S(i±^l  __•_]  =ffl)  Where,=  i. 

X  =  ocL  \  Xl\  Z  Jo         1+02  Jo  X 


434 


INTEGRAL  CALCULUS 


/Compare  \ 
e'     l  Art.  34.  / 


log(l+-)*]    =  zlog(l  +  j)]    =  a  (by  Ex.  6); 

-  H)1>-  -  K)1 

8.  (1  +  xY 1  =  e. 

(1+^=1-;  •••  ^(1+^=^^(1+4=^=1; 

.*.     (1  +  z)x    -  e.     (Compare  Cor.,  Art.  34.) 

9.  (1  -  a;)*]  =  -   or  e-K 

Jo       « 

iog(i-^]o=i.iog(i-*)]o=^yo=-i. 

.-.     (l-x)^  = 


e~x  or 


EXERCISE   XLV. 

Evaluate  the  following  indeterminate  forms: 


1  —  cos  x 
x1 


Jo      2" 


9. 


x-  1 


lJi     n 


3. 


tan  a:  — 


sin 
4.    (sinx)tar 


2.   sec  x  —  tan  x  L  =  0. 

-  sin  af|    _  1 
*X        Jo~2* 

Jr1- 

6.   zsin:cl    =  1. 

Jo 

6.  sin  x  log  z]o  =  0. 

7.  s*]o  =  1. 

8.  (l+x^]0  =  l. 


12.  £l^!l  =  2. 

sm.T    Jo 
a;  —  sin"1  .c~|         _  1 
sin3  x     Jo  6 

14.  tang-g-1   =  2 

a;  —  sin  x  Jo 

15.  (log  *)*]„  =  1. 

16.  aF^-e"1. 


EVALUATION  OF  DERIVATIVES 


435 


222.  Evaluation  of  Derivatives  of  Implicit  Functions.  — 
1.   Find  the  slope  of  a;4  -  a2xy  +  b2y2  =  0  at  (0,0).      Here 

?1    "TT^Vll     -J"      (Art-105)' 
cfojo.o      a2a;  -    2  622/Jo,o      0 

.2^ 


Hence      -/ 

ax  in.o 


dec 


12  x5 


da; 


2  6 


2*/ 
da; 


dx 


1.0 


a*-  26*^ 

dx. 


0,0 


...     * („,  _ 2  &*^U  «.  *]     =0;  whence  *]     =  0  or  £ 

2.   Find  the  slope  of  a;3  -  3  axy  +  y3  =  0  at  (0,0). 

Ans.   -rM     =  0  or   oo . 
aa;Jo,o 


CHAPTER  X. 

DIFFERENTIAL  EQUATIONS.     APPLICATIONS. 
CENTRAL   FORCES. 

223.  Differential  Equations.  —  A  differential  equation 
is  one  that  involves  one  or  more  differentials  or  derivatives. 
An  ordinary  differential  equation  is  a  differential  equation 
which  involves  one  independent  variable  only.  The  deriv- 
atives in  such  an  equation  are  therefore  ordinary  derivatives. 

The  order  of  a  differential  equation  is  the  order  of  the 
highest  differential  or  derivative  which  it  contains. 

The  degree  of  a  differential  equation  is  that  of  the  highest 
power  of  the  highest  differential  or  derivative  which  it  con- 
tains, after  the  equation  is  freed  from  fractions  and  radicals. 

For  examples : 

dy  =  mdx,  (1) 

are  ordinary  differential  equations.  Equation  (1)  is  of  the 
first  order  and  first  degree,  (2)  is  of  the  second  order  and  first 
degree,  and  (3)  is  of  the  second  order  and  second  degree, 
after  being  rationalized. 

224.  Solution  of  Differential  Equations.  —  It  has  been 
seen  in  foregoing  chapters  how  when  an  equation  expressing 
a  functional  relation  between  two  variables  is  given,  the 
differentiation  of  the  equation  gives  a  differential  equation 
expressing  the  rate  of  the  function.  On  the  other  hand,  it 
has  been  seen  that  the  rate  of  a  function  being  given  a  differ- 

436 


COMPLETE  INTEGRAL  437 

ential  equation  is  thereby  formed,  the  integration  of  which 
yields  the  function,  indeterminate  though  it  may  in  general 
be. 

It  is  this  rinding  of  the  function  from  an  equation  involving 
the  derivative  that  constitutes  the  solving  of  a  differential 
equation. 

The  general  solution  of  a  differential  equation  is  the  most 
general  equation  free  from  differentials  or  derivatives,  from 
which  the  given  equation  may  be  derived  by  differentiation. 

The  general  solution,  for  example,  of  the  equation 

g  =  C,  is  y  -  Cx  +  Ci, 

and  of  ■—  =  0,  is  y  =  Cx  +  Ch  also. 

dv  d2v 

Thus,  when  -~  represents  the  slope  and  -t~  the  flexion, 

this  function  is  any  non-vertical  line  in  the  plane.  Here, 
y  =  Cx,  y  =  Ci,  y  =  0,  y  =  x,  y  =  2  x  +  1,  .  .  .  are  par- 
ticular solutions,  which  are  included  in  the  general  solution. 
(See  Ex.  1,  Art.  115.) 

[Note.  —  The  general  solution  of  a  differential  equation 
may  not  include  all  possible  solutions.  A  solution  not  in- 
cluded in  the  general  solution  is  called  a  singular  solution. 
The  discussion  of  such  solutions  is  beyond  the  scope  of  this 
book.] 

The  general  solution  of  a  differential  equation  of  the  nth. 
order  contains  n  arbitrary  constants  of  integration  (for  ex- 
amples, see  Arts.  140, 141,  161);  to  determine  these  constants, 
n  conditions  connecting  the  function,  the  variable,  and  the 
successive  derivatives  must  be  known.  The  general  solution 
is  called  the  complete  integral  or  primitive  of  the  differential 
equation. 

225.  Complete  Integral.  —  When  a  differential  equation 
is  given,  passing  by  integration  to  the  complete  integral  is 


438  INTEGRAL  CALCULUS 

solving  the  equation.  It  has  been  proved  that  every  differ- 
ential equation  has  a  complete  integral,  and  that  when  the 
equation  is  of  the  nth  order  the  integral  contains  n  arbitrary 
constants.  The  complete  integral  then  contains  one,  two, 
...  or  n  constants  that  do  not  appear  in  the  differential 
equation,  when  that  equation  is  of  the  first,  second,  ...  or 
nth  order.  If  the  complete  integral  be  differentiated  these 
constants  are  eliminated,  whatever  may  be  their  particular 
values,  hence  they  are  called  arbitrary. 
Example  1.  —  Let  a  given  equation  be 

y  =  fp  +  k;  (i) 

dy      x 
differentiating,  -j-  =  - ;  (2) 

differentiating  again,     -^  =  -  =  -  ■£     (from  (2)) ;  (3) 

•••     *S-§  =  °>    «2)  Art- 223)  W 

is  the  differential  equation. 

To  find  a  general  form  for  the  complete  integral  of  (4), 

du  d?y      dm 

solve  by  letting  m  =  -~- ;    then  -=-*2  =  -7- ;   substituting  in 

...      .  dm  _  dm      dx 

(4) ,  gives  x  -; m  =  0,     or    —  =  — ; 

dx  m        x 

integrating,  log  m  =  log  x  +  log  c  =  log  ex, 

where  c  is  a  constant; 

dy  /rk,x 

.'.     m  =  ex,     or    -j-  =  cx\  (2 ) 

integrating  again  gives 

2/  =  !*2  +  C!,  (I') 

the  complete  integral,  in  which  c  and  C\  are  the  arbitrary 
constants;  and  which  has  the  form  of  equation  (1). 
Whatever  be  the  value  of  C\,  equation  (l7)  represents  a 


COMPLETE   INTEGRAL  439 

2 

parabola  on  the  y-axis  with  latus  rectum  - ;  hence  (2')  is  the 

c 

differential  equation  of  all  such  parabolas. 

Equation  (4)  is  the  differential  equation  of  all  parabolas 
whose  axes  are  on  the  y-Sixis. 

Suppose  (4)  is  given,  and  the  problem  is  to  find  a  function 
y  that  shall  satisfy  that  equation,  have  its  first  derivative 
equal  to  1/p  when  x  =  1,  and  be  equal  to  k  when  x  -  0. 
These  conditions  give,  from  (1')  and  (2'), 

fc  =  0  +  ci;     l/p  =  c, 

and  hence  the  function, 

y  =  x2/2  p  +  fc.     (Compare  Ex.  2,  Art.  161.)       (1) 

In  this  way  the  constants  can  be  determined  when  the 
necessary  conditions  are  known. 

N.B.  —  Equations  of  the  second  order  with  one  variable 

absent,  of  the  form  j '(yf ,  ~r,  x)  =  0,  may  often  be  solved  by  the 

method  used  in  this  example.     (Art.  232,  III.) 
Example  2.  —  Let  the  equation  be 

g  +  2  b  g  +  (V  +  <*)  y  =  0.     ((4)  Art.  54.)         (1) 

It  is  seen  in  Art.  54  that  this  differential  equation  results 
from  the  differentiation  of  the  equation 
y  =  ae~bt  sin  (ut  -a),    or    y  =  e~bt   (A  sin  ut  +  B  cos  ut) .     (2) 

To  show  that  (2)  is  the  solution  of  an  equation  of  the 
form  of  (1),  let  y  =  e~htu  and  (1)  becomes 

C^  +  ^u  =  0,  (3) 

where  co2  =  b2  +  co2  -  (i»2  6)2; 

then  by  Ex.  3,  following, 

u  =  A  sin  mt  +  B  cos  wt, 
and  y  =  e~bt(A  sin  ut  +  B  cos  ut).  (2') 

(Compare  Art.  233,  III.) 


440  INTEGRAL  CALCULUS 

Example  3.  —  Let  the  equation  be 

^|+co2u  =  0.     ((3)  of  Ex.  2.) 

Multiplying  by  2  du  gives 

integrating, 

/^Y  =  -^  (Ma  +  Ci)  =  co2  (a2  -  w2),     taking  ft  =  -a2, 

extracting  root,  -r-  =  to  Va2  —  w2. 

Integrating,  /  M       =  l  udt 

|/ 

gives  sin-1  -  =  co£  +  C2, 

or,  solving  for  w, 

w  =  a  sin  (co£  +  Cg)  =  A  sin  co£  +  5  cos  ut, 
where  A  =  a  cos  C2  and  5  =  a  sin  C2  are  arbitrary  constants. 
Example  4.  —  Let  the  equation  be 

Multiplying  by  2  du  gives 

/ciiA2 
integrating,  i-^j  =  co2(u2  +  Ci); 

extracting  root,  -rr  =  co  Vu2  +  Ci. 

Integrating,  /      .  =   /  co  eft, 

gives        log  (w  +  Vu2  +  Ci)  -  co£  +  C2, 

or,  solving  for  u,  u  =  Ae"1  +  £e^, 


THE   NEED  AND  FRUITFULNESS  441 

where  2  A  =  e°*  and  2B  =  —  de_C2  are  arbitrary  constants. 
By  means  of  the  hyperbolic  functions  this  result  may  be 
written  in  the  form 

u  =  a  sinh  (coZ)  +  b  cosh  (co£), 

where  b  +  a  =  2  A  and  b  —  a  =  2  B. 

Hence,  in  this  case,  a  solution  of  (1)  of  Ex.  2  is 

y  =  Ae-^-^1-^  Be-^+^K 
Example  5.  —  Let  the  equation  be 

^  =  0 
dt2      u' 

resulting  from  (3)  of  Ex.  2,  when  b2  +  w2  =  62,  or  co2  =  0. 

Integrating  gives   -77  =  &,  t*  =  Cit  +  C2. 

Hence  in  this  case,  the  solution  of  equation  (1)  of  Ex.  2,  is 
2,  =  e-u  [pxi  +  c2), 

where  C\  and  C2  are  constants. 

Note.  —  The  foregoing  examples  are  solutions  of  important 
differential  equations,  that  of  Ex.  2,  as  shown  in  Art.  54, 
being  the  typical  form  for  damped  vibrations. 

226.  The  Need  and  Fruitfulness  of  the  Solution  of 
Differential  Equations.  —  Attention  has  heretofore  been 
called  to  the  need  of  finding  the  inverse  of  a  rate,  in  solving 
many  problems  that  arise  in  everyday  life  as  well  as  in  science. 
In  fact  the  inverse  problem  is  more  often  the  real  question 
demanding  solution.  It  has  been  shown  (Ex.  5,  Art.  115) 
how,  when  the  acceleration,  the  rate  of  change  of  the  speed 
of  a  moving  body,  is  known,  the  velocity  and  the  distance  for 
any  time  are  found  by  the  solving  of  a  differential  equation. 
(Ex.  1,  Art.  161.) 

It  has  been  shown  (Art.  42),  that  when  a  function  has  the 
general  form  y  =  aebx,  the  rate  of  change  is  proportional  to 
the  function  itself,  and  that  so  many  changes  in  Nature 


442  INTEGRAL  CALCULUS 

occur  in  this  way  that  the  law  of  change,  known  as  the 
Compound  Interest  Law,  is  also  called  the  Law  of  Organic 
Growth.  Now,  if  it  is  known  that  some  function  changes  at 
a  rate  proportional  to  itself,  expressing  this  by  the  differen- 
tial equation, 

|  =  %,    or    kdx  =  df, 
then,      kx  =  f<^  =  \ogey  +  c,     or    y  =  ekx~c  =  Cekx, 

J   y 

where  C  =  e~c  is  an  arbitrary  constant. 

The  only  function  whose  rate  of  change  is  proportional  to 
itself  is  thus  shown  to  be  of  the  form  Cekx  (or  aebx),  where  C 
and  k  (or  a  and  b)  are  arbitrary,  and  k  (or  b)  is  the  factor 
of  proportionality.  This  may  be  expressed  also  by  the 
statement,  that  the  only  function,  whose  relative  rate  of 
change  (logarithmic  derivative)  is  constant,  is  Cekx  (or  aehx). 

It  has  been  shown  (Art.  73),  that  when  a  point  has  simple 
harmonic  motion  its  relative  acceleration  is  a  negative  con- 
stant. Thus,  when  the  displacement  of  a  point  is  given  by 
the  equation  y  =  a  sin  (cot  —  a),  there  results  the  differential 

d?u 
equation   —;  =  —o>2y,  where  co  is  constant,  and  hence  the 

relative  acceleration  is  -^  /  y  =  —  co2. 

Conversely,  when  the  motion,  as  in  a  vibration,  is  due  to 
a  force  that  increases  with  the  distance  from  the  central 
position,  the  acceleration,  being  according  to  Newton's 
second  law  of  motion  proportional  to  the  force,  is 

d2s  72 

at  =  de  =  ~ks> 

where,  as  the  force  acts  towards  the  origin,  the  acceleration 
is  negative  when  s  is  positive  and  positive  when  s  is  negative. 


THE  NEED  AND  FRUITFULNESS  443 

From  the  relation  v  dv  =  at  ds,  gotten  by  eliminating  dt  in 
dv/dt  =  at  and  ds/dt  =  v; 

J  vdv=  J  -k2s  ds;     :.     v2  =  d  -  k2s2; 

putting  Ci  =  k2a2, 

whence  sin-1 1-\  =  kt  +  C2 

or  s  =  a  sin  (kt  +  C2)  =  A  sin  kt  +  B  cos  kt, 

where  A  =  a  cos  C2  and  B  =  a  sin  C2  are  arbitrary  con- 
stants. This  equation  for  s  is  the  characteristic  equation  of 
simple  harmonic  motion;  the  amplitude  of  the  motion  is  a, 
the  period  is  2  w/k,  and  the  phase  is  —  Ci/k. 

Thus,  it  is  found  that,  when  the  acceleration  along  a 
straight  line  is  a  negative  constant  times  the  distance  from 
a  fixed  point,  the  only  motion  resulting  is  the  simple  har- 
monic motion. 

In  general,  it  has  been  shown  that,  whenever  the  rate  of 
change  of  a  function  of  a  single  independent  variable  is 
known  and  also  the  value  of  the  function  for  some  one  value 
of  the  variable,  it  is  possible  to  find  by  integration  the  value 
of  the  function  for  any  value  of  the  variable. 

Hence  it  has  followed  that  the  solution  of  a  differential 
equation  gives  the  area  under  any  curve  whose  equation  is 
known,  thus  solving  the  problem  that  had  baffled  the 
mathematicians  of  the  ages  before  the  discovery  of  this 
general  method  of  effecting  the  quadrature  of  curves  of  any 
degree. 

When  it  is  recalled  that  the  magnitude  of  any  quantity 
whatever,  whether  of  volume,  mass,  weight,  force,  work, 
etc.,  may  be  represented  by  an  area  under  a  curve,  the  fruit- 
fulness  of  the  solution  of  many  differential  equations  is 
recognized. 


444  INTEGRAL  CALCULUS 

Note.  —  In  this  chapter  and  in  the  foregoing  chapters  the 
differential  equations  that  have  been  solved  have  been  for 
the  most  part  ordinary,  involving  the  function  and  one 
independent  variable.  While  a  general  discussion  of  differ- 
ential equations,  including  those  other  than  ordinary,  is  too 
large  a  subject  for  a  first  course  in  the  Calculus  and  is  beyond 
the  scope  of  this  book,  some  of  the  special  equations  are  so 
important  that  their  solution  has  been  given,  and  some  more 
will  follow.  In  Art.  Ill,  the  solution  of  some  differential 
equations  that  have  the  form  M  dx  +  N  dy  =  0  was  effected, 
and  in  Art.  112,  the  definition  of  an  exact  differential  equa- 
tion of  that  form  was  given. 

227.  Equations  of  the  Form  Mdx  +  N  dy  =  0.  —  In  this 
form  M  and  N  are  functions  of  x  and  y.  The  variables  are 
said  to  be  separated  when  M,  or  the  coefficient  of  dx,  contains 
x  only,  and  N  contains  y  only. 

When  M  dx  +  N  dy  is  an  exact  differential  (as  defined 
in  Art.  Ill),  the  total  differential  of  some  function  of  x  and 
y,  then 

Mdx  +  Ndy  =  0,  (1) 

is  an  exact  differential  equation.    After  applying  the  test 

dM      dN      „a.  .   ,   111N 
-*y=Tx>     W)  Art.  Ill), 

and  finding  the  condition  satisfied,  integrate  the  coefficient 
of  dx  regarding  y  as  constant,  putting 

u=  fMdx+f(y),  (2) 

and  then  determining  /  (y)  so  that 

¥  =  N.  ■      (3) 

dy 

Or  regarding  x  first  as  constant,  put 

u  =  fNdy+f(x),  (4) 


EQUATIONS  445 


and  so  determine  /  (x)  that 

du 


dx  -  M-  <5> 

These  equations  involve  the  conditions, 

,r      du      %T      du  ,rt% 

M  =  d~x>    N  =  W  .  <6> 

Example.  —  Solve  (3  x2  +  4  xy)  dz  +  (2  z2  +  2  y)  cfy  =  0. 

— —  =  4-x  =  -^-,     hence,  the  condition  is  satisfied. 
ay  ax 

u  =  J{Zx2  +  ±xy)  dx  +f(y)  =  x*  +  2x2y  +f(y); 

^=2a*+f'(y)  =  2x*  +  2y, 

:.    f(y)=2y    and    f(y)=y2; 
hence 

u  =  x3  +  2  x2y  +  ?/2;       /.     x3  +  2  z2?/  +  ?/2  =  C 

is  a  solution  and  is  the  complete  integral. 
When  the  equation, 

M  dx  +  N  dy  =  0, 

is  not  exact,  it  may  be  multiplied  by  some  factor  that  will 
make  it  exact  in  some  cases.  This  factor  is  called  an  inte- 
grating factor.  Rules  have  been  given  for  finding  an  in- 
tegrating factor,  but  in  many  cases  a  factor,  or  several 
factors,  that  will  make  the  equation  become  exact,  may  be 
found  by  inspection.  For  Example,  see  Ex.  2,  Note,  Art.  Ill, 
y  dx  —  x  dy  =  0 

is  made  an  exact  differential  equation  by  either  the  factor 
£_2>  V~2,  or  (%y)~li  and  a  solution  effected  in  each  case. 
For  another  example, 

(1  +  xy)  y dx  +{1-  xy)  x dy  =  0, 
ydx-\-xdy  +  xy2  dx  —  x2y  dy  =  0, 
or  d  (xy)  -+-  xy2  dx  —  x2y  dy  =  0; 


446  INTEGRAL  CALCULUS 

dividing  by  x2y2  gives 

d{xy)      dx      dy  _  Q> 
(xy)2       x        y         ' 

1  x  1 

f-  log  -  =  log  c,     or    x  =  eye?*. 

xy  y 

228.  Variables  Separable.  —  An  equation  of  the  form 

Mdx+Ndy  =  0, 

in  which  the  variables  are  separated  can  be  solved  by  in- 
tegrating its  terms  separately.     The  variables  are  separable 
when  the  equation  can  be  put  in  the  form 
f(x)dx-+F(y)dy  =  0. 

Example  1.  —  cos  x  dx  —  sin  ydy  =  0. 
Here  sin  x  +  cos  y  =  C  is  evidently  the  general  solution. 

Example  2.  —  Va2  -  y2  dx  +  Va2  -  x2  dy  =  0. 

Dividing  by  Va2  -  y2  Va^-lc2 ;        .  dx       +     ,  dy       =  0 ; 

Va2  —  x2      Va2  —  y2 

x  n 

integrating,  sin-1  -  +  sin-1  -  =  C; 

Co  Q/ 

taking  sine  of  each  member,  the  first  member  being  a  sum, 
gives         x  Va2  —  y2  +  y  Va2  —  x2  =  a2  sin  C  =  C\. 
Example  3.  —  (1  —  x)  dy  —  (1  -f  y)  dx  =  0. 

Dividing  by    (1-*)(1+V);     =^L  -  ^  =  0; 

1+7/      1  —  x 

integrating, 

log  (1  +  y)  +  log  (1  -  x)  =  log  c  or  ch 
or  (1  +  ?/)  (1  —  x)  =  c  or  ^r>. 

The  final  equation  may  be  gotten  at  once  by  inspection. 

229.  Equations  Homogeneous  in  x  and  y.  —  If  an  equa- 
tion is  homogeneous  in  x  and  y,  the  substitution  of 

y  =  vx 


EQUATIONS  HOMOGENEOUS  IN  X  AND  Y         447 

fVT.ll  give  a  differential  equation  in  which  the  variables  are 
easily  separable. 

Example  1.  —  Solve  (x2  +  y2)  dx  —  2  xy  dy  =  0. 

Putting  y  —  vx  and  dividing  by  x2  gives 

(1  +  v2)  dx  -  2  v  (x  dv  +  v  dx)  =  0; 

separating  the  variables, _   2  =  0; 

ntegrating,  log  [x  (1  —  v2)]  =  logc; 

Dutting  y/x  for  y,  the  solution  becomes 
x2  —  y2  =  ex. 
Example  2.  —  Solve  (x2  +  y2)  dy  —  xy  dx  =  0. 
Putting  y  =  vx  and  dividing  by  z2  gives 
(1  +  v2)  (x  dv  +  v  dx)  —  v  dx  =  0; 

separating  the  variables,    1 (-  -=■  =  0 ; 

v        x       v3 

ntegrating,  log  v  +  log  x  —  §  v~2  =  C  =  log  c ; 

y  y       x2 

3utting  -  for  v  gives       log  -  =  7—^  > 

x  c       _  y~ 


)r  y  —  ce22/\ 

Example  3.  —  Find  the  system  of  curves  at  any  point  of 

;vhich  as  (x,  y),  the  subtangent  is  equal  to  the  sum  of  x  and  y. 

'  dx 

From  the  conditions,  the  subtangent  being  y  -z-  > 

dx 

lence,  ydx  —  xdy  =  y  dy; 

,.   .  ,.      ,       0     ydx  —  xdy      dy 
dividing  by  ?/2,     * ^ lf; 

x 
ntegrating,     -  =  log  y  +  log  Ci  =  log  dy, 

X 

Dr  ?/  =  celJ 

is  the  general  equation  of  the  system  of  curves. 


448  INTEGRAL  CALCULUS 

230.  Linear  Equations  of  the  First  Order.  —  A  linear 
differential  equation  is  one  in  which  the  dependent  variable 
and  its  differentials  appear  only  in  the  first  degree. 

The  form  of  the  linear  equation  of  the  first  order  is 

dy  +  Pydx  =  Qdx,  (1) 

where  P  and  Q  are  functions  of  x  or  are  constants. 

The  linear  equation  occurs  very  frequently.     The  solution 

of  dy  +  Pydx  =  0,     or    dy/y  +  P  dx  =  0, 

,  .     ,  fPdx  t  fPdx 

is  log  y  +  log  eJ        =  log  c,      or    yeJ        =  c. 

Differentiating  the  latter  form  gives 

efPdx{dy  +  Pydx)  =  0, 

/P  dx 
is  an  integrating  factor  of  (1).    Mul- 
tiplying (1)  by  this  factor  gives 

eJ    fJ (dy  +  Py  dx)  =  eJ     xQdx; 
and  this,  on  integration,  gives 

yefPdx=fefP,<*Qdx.  (2) 

The  equality  expressed  in  (2)  may  be  used  as  a  formula 
for  solving  any  linear  equation  in  the  general  form  (1). 
Example  1 .  —  Solve  xdy  —  ydx  —  x3  dx  =  0. 
Putting  it  in  the  general  form  (1),  it  becomes 

v 
dy  —  -dx  =  x2  dx. 
u      x 

Hence,  I  Pdx  =  —  I  —  =—  log  x  =  log  - ; 

fPdx  log]         1 


x 


and  /  eJ  PdxQ  dx  =  I  xdx  =  ^-  +  C. 


EQUATIONS  OF  ORDERS  ABOVE  THE  FIRST       449 

Substituting  these  values  in  formula  (2)  gives 
y-  =  \x2  +  C,    or    y  =  \x*  +  Cx. 

XL  J 

Example  2.  —  Solve  dy  +  y  dx  =  e^dx.      y  =  (x  +  C)  e-*. 

Example  3.  —  Solve  cos  x  •  dy  +  y  sin  x  •  dx  =  dx. 

y  =  sin  x  +  C  cos  x. 

Example  4.  —  Solve  (1  +  x2)  dy  —  yxdx  =  adx. 

y  =  ax  +  C  Vl  +  x2. 

Ti  a 

Example  5.  —  Solve  dy  -\--ydx  =  —dx.     xny  =  ax-\-C. 

231.  Equations  of  the  First  Order  and  nth  Degree.  —  An 

equation  of  the  first  order  and  nth  degree,  which  is  resolvable 
into  n  equivalent  rational  equations  of  the  first  degree,  may 
be  solved  by  the  solution  of  the  equivalent  equations.     To 

illustrate,  let  p  =  ~  in  the  examples. 

Example  1.  —  Solve  (W  +  (x  +  y)  ^  +  xy  =  0. 

The  given  equation  is  p2  +  (x  +  y)  p  +  xy  =  0;  factoring, 
(P  +  y)  (P  +  #)  =  0>  which  is  equivalent  to  the  equations, 
p  -f  7/  =  0,  p  +  £  =  0,  of  which  the  solutions  are, 
log  y  +  x  +  C  =  0,     2  y  +  x2  +  2  C  =  0. 
The  combined  solution  is 

(\ogy  +  x  +  C)(2y  +  x2  +  2C)  =  0. 
Example  2.  —  p2  -  ar3  =  0.  25  (y  +  C)2  =  4  ax5. 

Example  3.  —  p2  —  5  p  +  6  =  0. 

(y-2x-C)(y-Sx-C)  =  0. 
Example  4.  —  p3  -  ax4  =  0.  343  (y  +  C)3  =  27  a7. 

Example  5.  —  p3  +  2  xp2  -  y2p2  -  2  xy2p  =  0. 

(2/  -  C)  (y  +  x2  -  C)  (xy  +  Cy  +  1)  =  0. 

232.  Equations  of  Orders  above  the  First.  —  Examples 
will  be  given  of  solving  four  special  forms  of  such  equations. 


450  INTEGRAL  CALCULUS 

dny 

I.  Equations  of  the  form  j—n  =  f(x). 

The  solutions  of  equations  of  this  type  can  be  gotten  by 
n  successive  integrations.  Examples  have  already  been 
given  in  Art.  161  and  in  other  Articles. 

Example.  —  d4y  =  xz  dx\    y  =  ^  H — ^-  H — ^-  +  C3x  +  C4. 

d2y 

II.  Equations  of  the  form  -j-^  =  }{y). 

For  these  equations  2  dy  is  an  integrating  factor. 

Example.  —  Solve   ^  +  a2y  =  0.  (1) 

Multiplying  by  2  dy  gives 

integrating, 

^V  =  _ ay  +  Ci  =  a2  (Cl2  -  ?/2),     where  d  =  aV-     (2) 

From  (2),                 dy/VCl2  -  y2  =  adx;  (3) 

integrating  (3)                sin-1  ?//ci  =  ax  +  C2,  (4) 

or                                y  =  Ci  sin  (az  +  C2),  (5) 
which  may  be  written, 

y  =  A  sin  ax  +  £  cos  ax.  (6) 

See  Ex.  3,  Art.  225,  where  this  solution  was  obtained,  and 
in  Art.  226,  the  equation  for  simple  harmonic  motion  results; 
and  there,  for  the  differential  equation, 
d2s  _      M 
dt2  "      K  S> 
2  ds  might  be  used  as  an  integrating  factor. 

(dny  dy      \ 

-r-^,  .  .  .  ,   -j-,  xj  =  0;   that 

is,  equations  of  the  nth  order  with  y  absent. 


EQUATIONS  OF  ORDERS  ABOVE  THE  FIRST       451 

The  solution  of  a  differential  equation  of  this  form  was 
given  in  Ex.  1,  Art.  225. 

r>  4.  dy .     4-u        d2y      dp  dny      dn~1p 

Put  p  =  -r-,     then   ji  =  t1,  .  .  •  ,  -7-t  =  -3 — t  ■ 
•        da;  dx2      dx  dxn       dxn~l 

Substituting  these  values  in  the  general  form  gives 

/(£?.•■•.&»«)-*  (1) 

which  is  an  equation  of  the  (n  —  l)th  order  between  p  and  x. 
Example  1 .  —  To  show  that  the  circle  is  the  curve  for 
which  the  expression  for  radius  of  curvature  is  constant. 


? 


\dx,    ,    =  & 


dx2 

Substituting  p  for  dy/dx,  inverting  the  fractions,  and  sepa- 
rating the  variables,  gives 

dp        _  dx  < 
(l+p2)*~  #' 

integrating,  v        =  ±^A 

V 1  +  p  K 

a  being  arbitrary  constant; 
solving  for  p,  p  - 


dx  VR2-  (x-a)2' 

whence  y  —  b  =  ±  VR2  —  (x  —  a)2, 

b  being  arbitrary  constant;  hence,  (x  —  a)2  +  (y  —  b)2  =  R2, 
for  all  circles  of  radius  R. 

When  n  is  2,  the  equation  being  of  the  second  order,  the 
substitution 

dy       dp  =  fy 
V      dx'     dx      dx2' 
reduces  the  equation  given  to  an  equation  of  the  first  order 
in  dp/dx,  p,  x.     Solving,  if  possible,  gives  a  relation  of  the 


452  INTEGRAL  CALCULUS 

form  /  (p,  x,  C)  =  0.     This  is  still  of  the  first  order,  in  x  and 
y,  and  may  be  integrated. 

d2s      1       1 
Example  2.  —  Solve     -^  +  -r  +  ^  =  0. 

_,,.  ds     .         dp       1       1 

Putting  p  =  ^  gives  _  +  7  +  -  =  0. 

Separating  variables,  —  dp  =  —^ — dt; 

integrating,  —  p  =  —  -  +  log  t  +  Ci. 

Integrating  again, 

s  =  log*-  t\ogt  +  (l  -COt-  C2, 

which  gives  in  the  case  of  motion  the  relation  between  the 
space  or  distance  and  the  time. 

(dny\  dy      \ 

dW'  *  *  '  '   dx'  V)  =  °* 

Put   v-^'   then  *y-v&,    tl-tftE  +  JW?   etc 
rut   V-  dx,   then  dx2  -pdy}    dxZ  -  p  dy2  +  p ^j  ,  etc. 

Substituting  these  values  in  the  general  form  gives  an 
equation  of  the  (n  —  l)th  order  between  p  and  y. 

Thus  when  n  is  2,  the  equation  being  of  the  second  order, 
the  substitution 

_  dy         dp  _  d2y 
dx '        d?/      dx2 

reduces  the  given  equation  to  an  equation  of  the  first  order 

in  y  and  p.     This  is  solved,  if  possible;  and  then  dy/dx  put 

for  p,  giving  an  equation  in  x  and  y  of  the  first  order  to  be 

integrated  for  y. 

d2s 
Example.  —  Solve  at  =  -p  =  f(s). 

-n  i.  rfs    .,        d2s      vdv       .,  s 

Put  P-«-5.  then  ^-.gj. -/(.), 

giving  the  known  relation  vdv  =  at ds. 


LINEAR  EQUATIONS  OF  THE  SECOND  ORDER     453 

Integrating  gives  -  =  I  f(s)ds  +  C,  the  energy  integral, 
called  so  from  the  relation 

Fs  =  \mv2y 

the  equation  of  kinetic  energy,  where  Fs  is  work  done  by  a 
force  F  through  a  distance  s. 

When  /  0)  is  given,  v  is  replaced  by  ds/dt  and  the  integra- 
tion of  the  resulting  equation  gives  the  solution  in  terms  of  s 
and  t  and  the  equation  may  be  solved  for  s. 

Special  examples  under  this  case  were  given  in  II  and  in 
Art.  225,  Exs.  3  and  4. 

233.  Linear  Equations  of  the  Second  Order.  —  The 
general  form  of  the  equation  of  the  second  order  is 

where  P,  Q,  R  are  functions  of  x  alone  or  constants. 

The  complete  integral  of  all  linear  equations  is  the  sum  of 
two  functions,  called  the  complementary  function  and  the 
particular  integral. 

The  complementary  function  is  the  complete  integral  of 
the  equation  when  R,  the  term  independent  of  y  and  its 
derivatives,  is  zero.  This  function  will  contain  two  arbitrary 
constants,  when  the  equation  is  of  the  second  order. 

The  particular  integral  is  any  solution  of  the  equation  as 
it  is  in  the  general  form,  and  contains  no  arbitrary  constant. 

To  consider  the  complementary  function  in  which  P,  Q 
are  constants,  let  the  equation  be 

I.   Let  y  =  ekx,  k  constant;  then  substituting  in  (1)  gives 

(k2  +  ak  +  b)  ek*  =  0. 
If  'k  is  a  root  of  the  quadratic  equation, 

fc2  +  ak  +  b  =  0,  (2) 


454  INTEGRAL  CALCULUS 

called  the  auxiliary  equation,  ekx  will  satisfy  (1).     The  two 
roots  fci,  k2  of  (2)  are 

fcj  =  -ia  +  V|a2-  6,     fe2  =  -Ja-  Vj a2  -  6, 

and  efclX,  e*2*  are  two  solutions  of  (1).     Hence  the  complete 
integral  of  (1)  is 

y  =  £&*  +  Be^  =  erhax  (Aenx  +  5e"nx),  (3) 


where  u  =  v  J  a2  —  6.     For  special  cases: 

II.  If  a2  =  4  6,  equation  (2)  has  two  equal  roots,  k\  =  k2  = 
—  \  a.     In  this  case  (3)  becomes 

y  =  (A+£)e-*« 

where  (A  +  B)  might  be  replaced  by  one  constant  C.  When 
a2  =  4  b,  let  y  =  e~?axu  and  (1)  becomes,  without  the 
factor  e~*ax, 

^  =  0 
dx        ' 

of  which  the  complete  integral  is  u  =  A  +  Bx. 

Examples  of  this  solution  have  been  given  in  Arts.  224  and 
225.  The  complete  integral  of  (1)  when  the  auxiliary  equa- 
tion has  two  equal  roots,  each  —  \  a,  is 

y  =  (A  +  Bx)  e-*ax.  (4) 

III.  If  a2  <  4  b,  the  roots  of  (2)  are  imaginary.  Again, 
let  y  =  e~*ax  u  and  equation  (1)  becomes 

g  +  OTV=0,  (5) 

where  \a2  —  b  =  —  m2  and  m  is  real.  Now  (5)  is  satisfied 
by  it  =  cos  mx,  u  =  sin  mx;  its  complete  integral  is  then 

u  =  A  cos  mx  -f  B  sin  mx, 

and  therefore  the  complete  integral  of  (1),  when  a2  <  4  6,  is 

y  =  e-*axu  =  6_iax  (A  cos  mx  +  B  sin  wu).  (6) 

To  show  how  (3)  and  (6)  are  written  when  the  roots  of  (2) 


LINEAR  EQUATIONS  OF  THE  SECOND  ORDER     455 

are  known:  when  the  roots  of  (2)  are  real,  let  \  a2  —  b  =  n2; 
the  roots  are  then,  —  \a-\-n,—\a  —  n)  and  the  solution  is 

y  =  e-hax  (Aenx  _|_  Be~nx); 

when  the  roots  of  (2)  are  imaginary,  let  \  a2  —  b  =  —  n2, 
and  the  roots  are  then,  —\a-\-ni,  —  \a  —  ni;  and  the 
solution  is 

y  =  e-\ax  ^  cos  nx  -\-  B  sin  nx), 

so  that  instead  of  enix,  e~nix,  there  are  cos  nx,  sin  nx. 

It  may  be  noted  that  the  auxiliary  equation  is  written  by 

putting  k2  for  -y-| ,  k  for  -j- ,  and  omitting  y. 

The  solving  of  a  linear  equation  of  the  second  order  in- 
cluding these  three  cases  has  been  given  in  Art.  225,  Ex- 
amples 2,  3,4,5;  for  the  equation  for  damped  vibrations, 

Example  1.  —  Solve  §  +  8^  +  252/  =  0- 

The  auxiliary  equation  is 

k2  +  8  k  +  25  =  0, 

and  its  roots  are  h  =  —  4  +  3  i,  Ifa  —  —4  —  3 *.  Hence, 
the  complete  integral  is 

y  =  e~ix  (A  cos  3  x  +  B  sin  3  x) . 

Example  2.  —  Solve 

^-2^-35z/  =  0.     k2-  2k  -35  =  0. 
ax2        ax 

The  roots  of  the  auxiliary  equation  are 

ki  =  —5,     k2  =  7. 

Hence,  the  complete  integral  is 

y  =  Ae~bx  +  Be7x  =  ex  (Ae~6x  +  Be&x). 


456  INTEGRAL  CALCULUS 

APPLICATIONS. 

234.   Rectilinear  Motion.  —  I.   When  the  acceleration  is 
constant, 

da      d?s      n     d2s  ds 

s  =  i  at2  +  vot  +  s0. 

For  bodies  falling  freely  towards  the  earth  from  moderate 
heights,  the  acceleration  g  being  taken  constant, 

v  =  v0  -  gt,    s  =  tfct  —  |  gt2  +  «o; 
from  rest, 

v=  -gt,    s=  -\gt2.     (Ex.  5,  Art.  115.)  ' 

Projected  outward  from  rest,  h  =  v0t  —  \  gt2   (Ex.  6,  Art. 

116),  where  h  is  height  from  point  of  projection. 

II.   When  the  acceleration  varies  as  the  distance.     Let 

d2s 
a  =  -«  =  —  ks,  where  k,  a  constant,  is  the  acceleration  at 

a  unit's  distance  from  the  origin ;  and  let  the  body  be  of  unit 
mass  at  an  initial  distance  r;  then, 


*-(*)'- 


C\  —  ks2  =  kr2  —  ks2,  where  kr2  =  C\\ 


t  =  k~i  f    ,ds       =  h~+  cos"1  -  (  +  C2  -  0), 
J  Vr2  -  s2  r 


s 

£  =  0,  when  s  =  r; 
whence 


r  cos 


(kh)  =  r  sin  (kh+^\ 


where  the  last  result  is  gotten  if  the  positive  sign  of  the  radical 
is  taken  in  integrating. 


RECTILINEAR  MOTION  457 

Putting  s  =  0,  gives  v  -  r  Vk,  the  velocity  at  the  origin, 
and  t  =  7r/2  fc~a,  f  irk'*,  f  wk~*,  .  .  .  ,  or  s  =  r,  gives  t  =  0, 
27I-/&2,  4  7i7&2  ,  .  .  .  Hence  the  motion  is  periodic,  the 
period  being  2  *•/&»,  which  is  independent  of  the  initial 
distance. 

Differentiating  s  =  r  cos  (h*t),  gives 

v  =  jT  =  —  A^r  sin  (&^) ; 

which  expresses  the  velocity  in  terms  of  the  time  that  the 
body  has  been  moving. 

It  is  seen  that  this  is  a  case  of  simple  harmonic  motion. 
(Arts.  73  and  226.) 

It  has  been  shown  in  Art.  190,  Cor.,  that  the  attraction  of 
a  sphere  of  uniform  density  for  an  internal  particle  varies  as 
the  distance  from  the  center.  Hence,  a  particular  case  of 
the  periodic  motion  just  considered  would  be  that  of  a  body 
which  could  pass  freely  through  the  earth,  taken  as  a  homo- 
geneous sphere.  Such  a  body  would  vibrate  through  the 
center  from  surface  to  surface.  To  find  the  time  of  this 
half  period : 

t  =  Trfc-*  =  3.1416  V20900000/32.17  sec. 
=  42  min.  about. 

III.  When  the  acceleration  varies  inversely  as  the  square 
of  the  distance. 

Let  k  be  the.  acceleration  at  unit  distance  from  origin; 

..  d?s  k 

then,  „  =  _=_-, 

where  s  is  to  be  taken  always  positive;  multiplying  by  2  ds, 

2@)<:)=-2fo-2ds; 

integrating,  *  =  (|J  =  2k  g  -f),  (1) 


458  INTEGRAL  CALCULUS 

where  s  =  r,  when  v  =  0,  which  gives  the  velocity  of  a 
particle  at  any  distance  s. 


For  the  time 


/ 


2fc  —sds 


Vrs  —  s2 

negative,  since  s  decreases  as  t  increases, 

_  [1    r-  2s     _  r__J- I  ,  . 

|_2  y/rs  —  s2      2  Vrs  —  s2J 

integrating    between    limits    corresponding   to    t  =  t    and 
t  =  0,  gives 


=  V^[Vrs-s2-^vers~1T+?} 


(2) 


When  the  particle  arrives  at  the  origin,  s  =  0,  therefore,  the 
time  to  the  origin  from  the  point  where  s  =  r  is 


/r\i 


1      Vk\2 

It  is  seen  from  (1)  that  the  velocity  =  0  when  s  =  r,  and 
=  oo  when  s  =  0;  hence  the  particle  approaches  the  origin 
with  increasing  velocity.  While  the  attractive  force  causing 
the  acceleration  is  very  great  near  the  origin,  there  can  be 
no  attraction  at  the  origin  itself;  therefore,  the  particle 
must  pass  through  the  origin;  and  the  conditions  being  the 
same  on  either  side  of  the  origin  the  motion  must  be  retarded 
as  rapidly  as  it  was  accelerated;  hence,  the  particle  will  go 
to  a  point  at  a  distance  r  equal  to  that  from  which  it  started 
and  the  motion  will  continue  oscillatory. 

An  illustration  of  this  general  case  has  been  given  in 
Art.  193,  where  the  attraction  of  the  Earth  for  an  external 
particle  was  considered  as  the  cause  of  motion. 

IV.  When  the  acceleration  varies  as  the  distance  and  the 
motion  is  away  from  the  origin. 


RECTILINEAR   MOTION  459 

Let  k  again  be  the  acceleration  per   unit   mass  at  unit 
distance  from  the  origin;  then 

d2s      . 

a  =  de  =  ks'> 

using  2  ds  as  an  integrating  factor  gives 
2jLs.d(^)  =  2ksdS; 

/ds\2 
integrating,  v2  =  1-=-)  =  ks2  +  v02, 

where  v0  is  the  initial  velocity;  whence, 


s  = 


2  k 


(ekh  _  e-kh)m     (See  Ex  4?  Art  225.) 


Here,  as  t  increases  s  also  increases,  and  the  particle 
recedes  further  and  further  from  the  origin;  and  the  velocity 
also  increases  and  becomes  oo  when  s  =  t  =  oo.  Thus  in 
this  case  the  motion  is  not  oscillatory. 

V.  When  the  acceleration  is  constant  and  the  motion  is 
in  a  medium  whose  resistance  varies  as  the  square  of  the 
velocity. 

In  Art.  195  the  case  where  the  motion  was  towards  the 
Earth  has  been  given.  Let  now  the  particle  be  projected 
outward  with  a  given  velocity  v0.  Using  again  gk2,  as  the 
coefficient  of  resistance,  the  resistance  of  the  air  on  a  particle 
for  a  unit  of  velocity,  and  taking  the  particle  of  unit  mass, 
with  g  constant, 


('9 


whence,  '   ;  v>  =  —kgdtj 

integrating,       tan-1  (A-  --J  =  tan-1  (kv0)  —  kgt, 


460 


INTEGRAL  CALCULUS 


where  C  =  tan-1  (kv0) ;  solving, 
ds      1     kvQ 


v  = 


tan  kgt 


dt      k    1  +  kv0  tan  kgV  ^  ' 

which  gives  the  velocity  in  terms  of  the  time.     To  get  it  in 
terms  of  the  space ;  from  ( 1 ) , 


integrating, 


log 


1  +  kW 
where  c  =  log  (1  +  k2v02);  whence, 


=  -2gk2ds; 


-2gk2s, 


'-sr- 


=   n     =  v0'e- 


id 


e-2gk*s\ 


Writing  tan  %Z  in  (2)  in  terms  of  sine  and  cosine  and 
integrating, 

s  =  r—  log  (kv0  sin  kgt  +  cos  kgt), 
k  g 

which  gives  the  space  described  by  the  particle  in  terms  of 

the  time. 

235.   Curvilinear  Motion.  —  Let  a  body  slide  without 

friction  down  any  curve  ab.  The 

acceleration  caused  by  gravity 

at  any  point  P  is  g  sin  a,  where 

a  =  PTD,  PT  being  a  tangent 

to  the  curve. 

Let  PT  =  ds;   then  -PD  = 

dy;  hence, 

d2s  .  dy         ... 

-  =  gSlna=-g-.         (1) 

Let  y0  be  the  ordinate  of  the  initial  point  on  the  curve;  then 
v  =  0  when  y  =  y0. 


SIMPLE  CIRCULAR  PENDULUM 


461 


Integrating  (1)  gives 


v=jt  =  ^2g(yo-  y)- 


(2) 


It  follows  from  (2)  that  the  velocity  of  a  body  acquired 
by  moving  freely  down  any  frictionless  path  is  the  same, 
and  is  what  it  would  acquire  in  falling  freely  through  the 
vertical   height   between   the   initial   and   terminal   points. 

— ,  the  time  will  depend  upon  the  path. 

236.  Simple  Circular  Pendulum.  —  Consider  the  motion 
of  a  particle  on  a  smooth  circular  arc  under  the  action 
of  gravity  as  the  only  force. 


Taking  the  axis  of  x  as  vertical,  the  equation  of  the  circle  is 
y2  =  2  ax  -  x2.  (1) 

Let  K  (h,k)  be  the  point  where  the  particle  starts  from  rest, 
and  P  (x,  y)  where  it  is  at  the  time  t.  Then  the  particle  will 
have  fallen  through  the  height  h  —  x,  and  hence  from  (2), 
Art.  235, 

v  =  ft  =  V2g(h-x).  (2) 

It  is  seen  from  (2)  that  the  velocity  is  a  maximum  when 
x  =  0  and  a  minimum  when  x  =  h;  so  the  particle  will  pass 


462  INTEGRAL  CALCULUS 

through  0  following  the  curve  to  the  point  K'  where  x  =  h 
and  will  oscillate  between  K  and  K'. 

To  find  the  time  in  passing  from  K  to  K' )  from  (2), 

ii  _  —  -,  negative  since  s  decreases  as  t  increases, 

dt~  V2g(h-x) 

—  adx  ,         /-.N  j        adx 

:,  from  (1)  ds  =  —  •  (3) 


V2g{h-  x)  (2  ax  -  x2) 

While  this  expression  does  not  admit  of  direct  integration, 
approximate  values  of  the  integral  can  be  gotten  as  shown 
in  Ex.  6,  Art.  203.  When  the  arc  is  small,  the  approximate 
value  of  the  time  can  be  gotten  from  (3)  thus : 

T  =  2*=  2      P°  adx 


gJh 


V2gJh  V(h-x)x(2a-x) 
h       dx 


Vh 


x  —  X1 


by  taking  (2  a  —  x)  =  2  a, 


-v&-*¥I-'\/r  » 

If  instead  of  moving  on  a  curve,  the  particle  is  assumed  to 
be  suspended  by  a  rod  or  cord  of  no  weight,  it  becomes  a 
simple  pendulum,  existing  in  theory. 

To  reduce  (3)  to  the  form  known  as  "an  elliptic  integral" 
of  the  first  kind;  let  h  =  a  vers  a,  x  =  a  vers  6,  a  =  KCO, 
being  constant  and  6  variable ;    then 

h  —  x  =  a  (vers  a  —  vers  0)  =  a  (cos  6  —  cos  a) ; 
dx  =  a  sin  0  dd;    and     (2  ax  —  x2)  =  a2  sin2  6. 

Substituting  in  (3) : 

2  a  •  a  sin  6  dd 


r=-i-r_ 

V2gJ0   y/a 


(cos  6  —  cos  a)  a2  sin2  0 
dd 


gJo   Vsin2a/2  -  sin2  0/2 


SIMPLE  CIRCULAR  PENDULUM  463 


-v/*P 


d$ 


Jo  2  V  Vsina/2/ 

2  sin  a/2  cos  <f>  d<f> 


= Ji  r 

v  0  J0  sin  a/2  Vl  -sin2<£  Vl  -  sin2  a/2  sin2  0 

=  2  V  -  /       ,  .,  by  cancellation.      (5) 

V  9  J0  Vl  -sin2  a/2  sin2  0 

Here  <f>  is  denned  by  sin  <j>  =  - — 4r,  giving  by  differenti- 

sin  a/  _ 

ation, 

4jJ       cos  B/2de/2 

COS0C?0  =  — r1 — 7?p— , 

sm  a/2 

whence, 

J,,  _  2  sin  a/2  cos  <f>  d<f>  _     2  sin  a  /2  cos  0  d<£ 
cos  0/2  "  Vl  -  sin2  a/2  sin2  <j> 

For  change  of  limits,   6  =  a  when  0  =  7r/2,   and  0  =  0 
when  0  =  0. 

Putting  k  =  sin  a/2  in  (5),  it  becomes 


,—     * 
T=*2\/-   T(l  -  k2  sin2  0)"*d0 

=  2  V^X"  (*  +  ^'2  Si"2  0  +  l'Ifc4  Si"4  0  +  '  '  '  )d* 

"^h@'^(H)V+(liI)V+.}<6, 

?r 

The  form    /     (1  —  k2  sin2  <£)~^  d<fr  is  "aw  elliptic  integral 

of  the  first  kind."  . 

When  a,  and  hence  k,  is  small,  only  the  first  two  terms  of 
the  final  series  will  give  a  close  approximation  to  the  value 

of  T,  and  the  first  term  alone  gives  the  value  ir  y  - ,  the  expres- 


464 


INTEGRAL  CALCULUS 


sion  usually  taken  as  the  time  of  a  vibration;  2icy  -  being 

the  value  taken  for  a  complete  oscillation  back  and  forth. 

237.   Cycloidal  Pendulum.  —  A  particle  moves  along  the 
arc  of  a  cycloid;  find  the  time  of  descent. 


From  (2)  of  Art.  235, 


ds 


v  =  Jt  =  ^2g(y0-y). 
The  equation  of  the  cycloid  referred  to  OX  and  OY  is 

x  =  a  vers-1  y/a  +  V2  ay  —  y2. 
Hence,  ds  =  —V2a/y  dy,  which  substituted  in  (1),  gives 


(1) 


t 


la  p_^=  =  v/^vers.12j,>=  JaY        ^M 

V  gJv    Vvny-V2       V  a  VoJv        V  0L  2/oJ 


t 


vyQy-y2      *  Q  V» 

Try-,    when    y  =  0,    and    2t=T 


9 


the  time  of  one  oscillation  of  a  pendulum  if  it  swings  in  the 
arc  of  a  cycloid. 

The  time  of  an  oscillation  being  independent  of  the  length 
of  the  arc,  the  cycloidal  pendulum  is  isochronal. 

The  pendulum  is  described  in  Art.  97,  Ex.  3. 

The  cycloid  is  the  curve  of  quickest  descent,  the  Brachysto- 
chrone;  that  is,  the  curve  down  which  without  friction 
gravity  will  cause  a  particle  to  fall  in  the  shortest  time. 


CYCLOIDAL   PENDULUM 


465 


The  following  is  a  comparison  between  the  times  down  the 
chord  of  a  circular  arc,  a  circular  arc,  and  the  arc  of  a  cycloid. 
For  the  time  down  the  chord  I  of  a  circle 


y2  =  2  rx  —  x 


••  -Vj 


d2s  I 

From  (1),  Art.  235,  -p  =  gsina  =  g  — ,  from  the  circle, 

V  =  Jt='2rt  (+C  =  °'  since  y  =  °'  when  '  =  °^ 
S  =  hr  f~  (+C  =  °>  since  s  =  0j  when  l  =  0)t 

<=\/¥=2v/^whens=L 


c 

X 

^sT 

s 

B 

ta          \ 

(, 

J   J^^y^ 

V" 

For  the  time  down  the  circular  arc  AO; 

2  V  sf  L       4  V2  +  4/  ^  64  W2  +4/  ^         J   Urt.  236/ 

=  l\-  [1+0.07+0.01+  •  •  •  ]=0.54x\A-(1.7...)V^ 

Z  f   gr  f  g  V  g 

For  the  time  down  the  arc  of  a  cycloid  from  A  to  0; 


466 


INTEGRAL  CALCULUS 


From  the  figure, 


7T2+4 

r  =  — - — a 


or    a  = 


4r 


hence 


fc2 


7T2+4    ' 

or     I2  =  (2r-2a). 

V^=a-6...)^<(i.7...)V^<^- 


Hence, 

It  is  at  once  evident  that 


It  may  be  seen  that  the  approximate  value 

lv/^(i.57...)v/-;<a-6...)v/^ 

238.  The  Centrifugal  Railway.  —  The  centrifugal  rail- 
way is  an  example  of  a  simple  circular  pendulum  where  the 
cord  of  suspension  is  replaced  by  a  track. 

Neglecting  the  resistance  of 
friction  and  of  the  air,  the  forces 
acting  on  the  car  are  the  force  of 
gravity  and  the  normal  reaction 
of  the  track.  If  h'  is  the  diam- 
eter of  the  circular  track  and  h 
the  height  from  which  the  car 
starts  from  rest,  find  the  relation 
between  h  and  hf,  so  that  the  car 
will  make  a  complete  revolution  without  leaving  the  track. 
The  centrifugal  reaction  at  the  highest  point  of  the  track 
must  be  great  enough  to  overbalance  the  weight  of  the  car. 
The  velocity  at  B  is  v  =  V2gh;   at  T,  v  =  V2g  (h-  h') ; 

Wv2         W 
whence,  ^  =  -^  ■  2  g  (h  -  h')  =  W, 

giving  4    (h  —  h')  =  h';   hence  h  =  $  h',  to   balance;   and, 


PATH  OF  A  LIQUID  JET  467 

therefore,  h  should  be  greater  than  f  h' ,  for  the  car  to  com- 
plete the  revolution  without  leaving  the  track. 

239.  Path  of  a  Liquid  Jet.  —  If  a  small  orifice  be  made 
in  the  vertical  side  of  a  vessel  containing  a  liquid  like  water, 
and  a  short  tube  be  inserted  so  as  to  direct  the  current 
obliquely,  horizontally,  or  vertically  upward,  the  velocity  of 
efflux  will  be  the  same,  since  the  .pressure  of  fluids  at  the 
same  depth  is  the  same  in  every  direction.  To  find  this 
velocity,  let  v  be  the  velocity,  w  the  weight  of  the  liquid 
issuing  with  that  velocity  per  second,  and  h  the  head  or 
height  of  the  surface  of  the  liquid  above  the  orifice;  then 
the  equation  for  energy  is 

wh  =  7:  —  v2, 
2  g 

in  which  wh  is  the  work  w  can  do  in  falling  through  h,  and 

1  w 

-  —  v2  is  stored  up  energy  in  w  as  it  issues  from  the  orifice. 

^  9 

Supposing  no  loss  of  energy,  they  are  equal;  hence, 

v2  =  2gh     or     v  =  V2gh;  (1) 

that  is,  the  velocity  of  efflux  is  the  same  as  that  of  a  body  which 
has  fallen  freely  through  the  height  h. 

Now  each  particle  of  liquid  issuing  from  the  orifice  will 
have  the  same  velocity  and  will  follow  the  same  path.  The 
path,  when  the  tube  is  not  vertical,  will  be  a  parabola  whose 
directrix  is  fixed  in  the  surface  of  the  liquid  supposed  to  be 
kept  at  a  constant  level  by  more  liquid  entering  the  vessel 
(Art.  196).  If  the  liquid  issue  obliquely,  its  equation  is 
given  in  (3),  Art.  196.  If  the  liquid  issue  horizontally,  a  =  0, 
and  the  equation  becomes 

^  =  ^  =  4%.  (2) 

The  equation  (2)  may  be  derived  thus :  let  a  particle  issue 


468 


INTEGRAL  CALCULUS 


from  the  orifice  with  a  velocity  v,  and  in  t  seconds  be  at  a 

point  P;  then 

x  =  vt  (distance,  v  constant), 
V  =  igi2  (freely  falling  body). 

Eliminating  t  between  these  equations  gives 


h    — 


If  the  x  and  y  of  any  point  of  the  jet  is  measured,  the  equa- 
tion (2)  can  be  used  to  determine  the  actual  velocity  of  flow 
from  the  orifice.  If  this  is  done,  the  coefficient  of  velocity 
is  given  by 

actual  velocity 


cv  = 


theoretical  velocity 


the  actual  velocity  being  less  than  the  theoretical  on  account 
of  friction  at  the  edge  of  the  orifice. 

The  path  is  derived  without  taking  into  account  the 
resistance  of  the  air,  as  when  the  path  is  in  a  vacuum. 

When  the  orifice  is  at  the  center  of  the  vertical  side  of  a 
vessel  kept  constantly  full  of  liquid,  it  can  be  easily  shown 
that  the  horizontal  range  of  the  jet  is  a  maximum  and  equal 


DISCHARGE  FROM   AN  ORIFICE 


469 


to  2  h,  the  height  of  the  vessel;  and  at  equal  distances  above 
and  below  the  center  the  range  will  be  the  same. 

The  coefficient  of  velocity  for  a  small  sharp  edge  orifice  is 
0.98;  and,  for  a  short  tube,  it  is  about  0.82. 

240.  Discharge  from  an  Orifice.  —  If  a  is  the  area  ofa 
small  orifice,  then  the  theoretical  discharge,  or  the  quantity 
of  liquid  issuing  in  a  unit  of  time,  is 

Q  =  av  =  aV2~gh.  (1) 

On  account  of  the  contraction  of  the  jet  at  the  orifice  and  the 
diminution  of  the  velocity  the  actual  discharge  is 


Q  =  cccvav  =  0.6  a  V2  gh, 


(2) 


where  cd  =  cccv  =  0.62  X  0.98  =  0.6  about,  for  a  standard 
orifice;  for  a  standard  short  tube,  cd  =  0.82,  the  coefficient 
of  contraction  being  unity. 


For  a  small  orifice  the  head  is  taken  as  constant  and  as 
that  on  the  center,  and  for  heads  greater  than  twice  the 
height  of  the  orifice  that  gives  the  discharge  almost  exactly. 
For  large  orifices  under  low  heads  the  variation  of  head  over 
the  orifice,  causing  a  variation  in  the  velocity  of  the  jet  and 
therefore  in  the  discharge,  makes  the  formulas  above  in- 
applicable for  exact  results. 

Let  h  be  the  head  on  the  center  of  a  rectangular  orifice  of 
breadth  b  and  depth  d;   and  let  the  rectangle  be  supposed 


470  INTEGRAL  CALCULUS 

to  be  divided  into  horizontal  strips  of  area  b  Az,  x  being  the 
distance  from  the  center  line  of  the  rectangle.     The  quantity 

d 


Q  =  lim  yjblxV2g(h-  x)  =  bV2g  I     (h-x)^dx; 

Ax=0  ^  J _d 

2 

J   d\        2h      Sh2  J        \ing  in  series  / 

—  2 

=  MV2^(l-9-^-^gg--..).  (3) 

It  is  seen  that  the  quantity  in  the  parenthesis  is  less  than 
unity,  and  the  discharge  is  therefore  less  than  that  given 
by  (l). 

For  h  =  2  d,  the  value  of  the  parenthesis  factor  is  0.997, 
so  for  heads  greater  than  twice  the  height  of  the  rectangle 
the  discharge  may  be  figured  from  Q  =  caV2  gh,  where  c  is 
the  coefficient  of  discharge. 

Integrating  without  expanding  (h  —  x)*  gives 

d 

Q  =  bV2~g  C(h-x)Ux  =  bV2gl-  | (&-*)*[ 

-|wi;[(, +#-(,- ft 

If  the  orifice  extends  to  the  surface  and  the  bottom  is  h 
below, 


Q  =  bV2g[-i(h-  *)*T  =  1  bh  V2gh, 


(5) 


which  is  just  §  the  quantity  that  would  flow  through  an  orifice 
of  equal  area  placed  horizontally  at  the  depth  h,  the  vessel  being 
kept  constantly  full. 

The  mean  velocity  vm  is  seen  in  (5)  to  be  §  V2  gh. 

For  any  vertical  orifice  formed  by  a  plane  curve  whose 
vertex  0  is  at  the  depth  hi  below  the  surface  of  the  liquid  in 


DISCHARGE  FROM   AN  ORIFICE 


471 


a  vessel  of  height  h,  kept  constantly  full,  the  formula  for  dis- 
charge is 

Q=   f'~12yV2g(h1  +  x)dx.  (6) 

To  get  the  time  of  emptying  the  vessel;  let  the  surface  be 
z  below  the  top  at  the  end  of  the  time 
t,  z  —  0  when  t  =  0;  then  the  quantity 
discharged  in  an  element  of  time  is 


dQ  =  hV2g£  ^yVx  +  h 


zdx 


]dt, 


h,  4Z 


z  being  constant  during  this  integration; 

and  since  in  the  same  time  the  quantity 

discharged  through  the  orifice  must  be 

A  dz,  A  being  the  area  of  the  section  of  the  vessel  at  depth 

z,  it  follows  that 

\2  V2g  jy  Vx  +  h-z  dx\  dt  =  A  dz; 

1_    Ch Adz 

2  9  I      fh  h\,  y/x  +  h-z  dx 


t  = 


(7) 


Example.  —  Water  is  flowing  from  an  orifice  in  the  side 
of  a  cylindrical  tank  whose  cross  section  is  100  sq.  ft. 
The  velocity  of  the  jet  is  V2  gx,  x  being  the  height  of  the 
surface  above  the  orifice;  and  the  cross  section  of  the  jet  is 
0.01  sq.  ft.  Find  the  time  it  will  take  for  the  water  to  fall 
from  100  ft.  to  81  ft.  above  the  orifice. 

For  this  example  the  formula  becomes 

A        Chl  -i 

t  = -7=  I    x  2  dx   (where  x  is  height  of   surface 

ca  V2  g  Jh 

above  orifice  and  a  is  area  of  the  orifice) 

100=  PV*  dx=-  15200  §  2  hflL     /taking\ 

2  a  Jioo  8  Jioo    \g  =  32/ 


0.01  V2„ 
=  2500  (10  -  9)  =  2500  sec.  =  41f  min. 


472  INTEGRAL  CALCULUS 

CENTRAL  FORCES. 

241.  Definitions.  —  A  central  force  is  one  which  acts 
directly  towards  or  from  a  fixed  point  and  is  called  an  attrac- 
tive or  a  repulsive  force  according  as  its  action  on  any  particle 
is  attraction  or  repulsion.  The  fixed  point  is  called  the 
center.  The  intensity  of  the  force  is  some  function  of  its 
distance  from  the  center. 

The  path  of  the  particle  is  called  its  orbit.  All  the  forces 
of  Nature  that  are  known  are  central  forces. 

242.  Force  Variable  and  Not  in  the  Direction  of  Motion. 
—  Let  a  particle  of  unit  mass  be  projected  in  any  direction 

and  acted  on  by  an  attractive 
force  F.  The  path  will  be  in 
the  plane  passing  through  the 
center  of  force  and  the  line  of 
projection. 

In  this  plane  let  0,  the  center 
of  attraction,  be  the  origin  and 
the  pole,  and  let  (x,  y)  or  (p,  6)  be  the  position  of  the  par- 
ticle P  at  the  time  t.     The  equations  of  motion  are,  from 
the  components  of  F  parallel  to  the  axes  OX  and  OY, 

dL£=-Fc0se=-F-,    ^=  -Fsind=  -F^-       (1) 
dv  p      dt2  p 

Multiplying  the  first  by  y  and  the  second  by  x  and  sub- 
tracting, 

(3) 


integrating, 

dy         dx      , 

xdt-yTt  =  h' 

where  h  is  a 

constant. 

From 

x  = 
dx  = 
dy  = 

p  cos  0    and     y  =  p 
cos  6  dp  —  p  sin  6  d0, 
sin  6  dp  +  p  cos  6  dd, 

sin 

• 

(4) 


FORCE  VARIABLE 


473 


which  in  (3)  gives 


dO 

dt 


(5) 


Multiplying  equations  (1)  by  2  cfo  and  2  dy  and  adding: 
2dx(Px  +  2dyd2y  =    _  2F(xdx  +  ydy)  . 
dt2  p 

-  '((SHs)")-^ 

Putting  p  =  -   and   hence  dp  = ^,   (7)  becomes 


(6) 


(7) 


whence 


d?u         _  _F_ 


2F 


2dw, 


/i2U2 


0, 


(8) 


which  is  the  differential  equation  of  the  path;  and  as  the  force 
F  will  be  given  in  terms  of  p,  and  therefore  in  terms  of  u,  the 
integral  of  the  equation  will  be  the  polar  equation  of  the  path. 

Let  the  central  attraction  vary 
inversely  as  the  square  of  the  dis- 
tance; to  find  the  path. 

Let  the  particle  be  projected 
from  the  point  P0  with  a  velocity 
V,  R  the  value  of  p  for  the  point 
P0,  j8  the  angle  between  R  and  the 
line  of  projection;  and  let  K  be 
attraction  for  the  unit  mass  at  unit  distance,  and  t  =  0  when 
the  particle  is  projected.  Then  since  the  perpendicular  from 
the  origin  to  the  tangent  is 

p  =  p2  —   [Art.  777  (9)],  the  velocity   ~n  =  ~>  from  (5); 


474  INTEGRAL  CALCULUS 

and  as  at  P0,  p  =  R  sin  0, 

7=— t-;     h=VRsmp. 
Rsm(3 

As  the  force  varies  inversely  as  the  square  of  the  distance, 

TC  1 

F  =  -=■  =  Ku2,     where  u  =  - ; 
P2  P 

d2u  K 

hence,  ^-  +  u  =  -p,  from  (8).  (8') 

Using  2  dw  as  integrating  factor  and  integrating, 

when  I  -  0,   t*  =  -  =  ^  and    ^j  +^2=F  by  *  =  -2 

and  Art.  77  (9); 

T^2  _  2X  =  V2R-2K 
"  h2      h2R  h2R        ; 

substituting  this  gives 

»W  |U,_H-2JC     2fa 

Hence 

which  shows  that  the  velocity  is  greatest  when  p  is  least,  and 
least  when  p  is  greatest. 

Changing  the  form  of  (9)  to 

du2      V2R-2K  .  K2 


K2      IK        V 

+  TF-fc— )•  (10) 


dd2  h2R 


and  simplifying  by  letting  u  =  b,  and jjiy—  +  jl  =  <?> 


gives 


[c2  -  (u  -  &)•]« 


FORCE  VARIABLE  475 

the  negative  sign  of  the  radical  being  taken.  Integrating 
gives 

_t  u  —  b  , 

cos  1 =  6  —  c, 

c 

where  c'  is  an  arbitrary  constant; 

/.     u  =  6  +  ccos(0-  c').  (11) 

Replacing  in  (11)  the  values  of  b  and  c  and  the  value  of  h, 
and  dividing  both  terms  of  the  second  member  by  K,  gives 

=  1= WW/3/if 

P     u      l  +  [l/K2(V2R-2K)RV2sm2p+l]2cos(^-c'), 

which  is  the  polar  equation  of  the  path  and  is  the  equation 
of  a  conic  section,  the  pole  being  at  the  focus,  and  the  angle 
(6  —  c')  being  measured  from  the  shorter  length  of  the  major 
axis.  For  if  e  is  the  eccentricity  of  a  conic  section,  p  the 
focal  radius  vector,  and  <f>  the  angle  between  p  and  that  point 
of  a  conic  section  which  is  nearest  the  focus,  then 

1        a(l-e2)  a(e2-l)  /10, 

u       1  +  e  cos  <{>  1  +  e  cos  0 

Comparing  (12)  and  (13),  it  is  seen  that, 

e2  =  l/K2  (V2R  -2K)  KV2  sin2  0  +  1 ;  (14) 

<t>  =  e  -  c'.  (15) 

Since  the  conic  section  is  an  ellipse,  parabola,  or  hyper- 
bola, according  as  e  is  less  than,  equal  to,  or  greater  than 
unity,  and  from  (14),  e  is  thus,  according  as  V2R  —  2  K  is 
negative,  zero,  or  positive;  therefore,  it  is  seen  that 

V2  <  -5-,  e  <  1,  and  the  orbit  is  an  ellipse, 

2  K 
V2  =  -5-,  e=l,  and  the  orbit  is  a  parabola, 
it 
9  js 

V2  >  -5- ,  e  >  1,  and  the  orbit  is  a  hyperbola. 
K 


476  INTEGRAL  CALCULUS 

Corollary  1.  —  By  (1)  of  Art.  234,  III,  it  is  seen  that  the 
square  of  the  velocity  of  a  particle  falling  through  an  infinite 
distance  to  a  point  R  distant  from  the  center  of  force  is  under 
the  law  of  attraction  now  considered  2  K/R.  Hence  the 
conditions  above  may  be  expressed  by  stating  that  the  orbit, 
described  about  this  center  of  force,  will  be  an  ellipse,  a 
parabola,  or  a  hyperbola,  according  as  the  velocity  of  pro- 
jection is  less  than,  equal  to,  or  greater  than  the  velocity 
through  an  infinite  distance. 

Corollary  2.  —  From  (15)  it  is  seen  that  6  —  c'  is  the  angle 

between  the  focal  radius  vector  p,  and  that  part  of  the 

principal  axis  which  is  between  the  focus  and  the  point  of 

the  orbit  which  is  nearest  the  focus;   that  is,  it  is  the  angle 

POA  in  the  figure;   and  hence,  if  the  principal  axis  is  the 

initial  line,  c'  =  0. 

2  K 
Corollary  3.  —  If  the  orbit  is  an  ellipse,  V2  <  —^- ;   hence, 

K 

e2=l-  1/K2  (2K  -  V2R)  RV2sin2p,  from  (14).       (16) 
The  polar  equation  of  the  ellipse  is 

a  (1  -  e2)  . 


P  = 


1  +  e  cos  0 


comparing  it  with  (12),  corresponding  terms  give 

n        -       R2V2sm2(S. 

a  (I  -  e2)  = ^ , 

substituting  for  1  —  e2  its  value  from  (16)  and  solving  for  a; 

KR 


2K-  V2R' 


(17) 


which  shows  that  the  major  axis  is  independent  of  the  direction 
of  projection. 

In  the  figure,  P0  is  the  point  of  projection;  FP0  =  R;  P0T 
is  the  line  along  which  the  particle  is  projected  with  velocity 
V;  FPQT  =  j8,  the  angle  of  projection;  FP  =  p;  PFA  =  d; 


KEPLER'S  LAWS  OF  PLANETARY   MOTION         477 

FT  =  p  =  R  sin  &;  if  /3  =  90°,  the  particle  is  projected  from 
an  apse,  A  or  A',  an  end  of  the  major  axis. 

To  determine  the  apsidal  distances,  FA  and  FA';    -^  =  0; 

dd 

•'•  "2-¥  +  ll_fc2  =  0'from(9)'      (18) 

the  two  roots  of  which  are  the  reciprocals  of  the  two  apsidal 
distances,  a  (1  —  e)  and  a  (1  +  e). 

Since  the  coefficient  of  the  second  term  of  (18)  is  the  sum 
of  the  roots  with  their  signs  changed, 

1  1 2K        .         n        _       h2      nm 

+  „m    i   ^  =  "IT  i      •  •      «  (1  -  e2)  =  ^.     (19) 


a(l-e)  '  a(l+e)        /i2   7  v  J      K 

which  is  one-half  the  latus  rectum. 

From  (5)  p2  dd  =  h  dt;  and  area  swept  over  by  the  radius 
vector  is 

2A 
h 

which  shows  that  the  areas  swept  over  by  the  radius  vector  in 
different  times  are  proportional  to  the  times,  and  equal  areas 
will  be  described  in  equal  times.     If  t  =  1,  A  =  §  h\  hence, 
h  =  twice  the  sectorial  area  described  in  a  unit  of  time. 
For  the  time  of  describing  the  ellipse,  calling  the  time,  T; 


A  =  \J?d0  =  \£hdt  =  \ht;     .:     t  = 


T  _  2  area  of  ellipse  _  2 


TTOb 


2  7ra2  Vl  -  e 


h 
(from  (19)) 


VKa  (1  -  e2) 

which  is  the  periodic  time. 

243.  Kepler's  Laws  of  Planetary  Motion.  —  From  a  long 
series  of  observations  of  the  planets,  especially  of  Mars, 
Kepler  deduced  the  following  three  laws  which  completely 
describe  planetary  motion. 


478  INTEGRAL  CALCULUS 

I.  The  orbits  of  the  planets  are  ellipses,  of  which  the  sun 
occupies  a  focus. 

II.  The  radius  vector  of  each  planet  describes  equal  areas 
in  equal  times. 

III.  The  squares  of  the  periodic  times  of  the  planets  are 
as  the  cubes  of  the  major  axes  of  their  orbits. 

The  statement  of  these  laws  marked  an  epoch  in  the 
development  of  mechanics,  for  the  investigations  of  Newton 
as  to  the  nature  of  the  attractive  force  led  to  his  discovery  of 
the  law  of  universal  gravitation.  The  conclusions  deduced 
by  Newton  from  Kepler's  three  laws  will  be  briefly  shown. 

244.  Nature  of  the  Force  which  Acts  upon  the  Planets.  — 
(1)  From  the  second  of  Kepler's  Laws,  it  follows  that  the 

planets  are  retained  in  their  orbits 
by  an  attraction  tending  towards 
the  Sun. 

Let  (x,  y)  be  the  position  of  a 
planet  at  the  time  t,  referred  to 
rectangular  axes  through  the  Sun 
in  the  plane  of  the  motion  of 
the  planet;  X,  Y,  the  component 
accelerations  due  to  the  attraction  acting  on  it,  resolved 
parallel  to  the  axes;   then  the  equations  of  motion  are, 

^1  =  y      ^1  =  v 
dt2         '      dt2          ' 

■•■  xC3-ydi  =  xY-yX-  (1) 

By  Kepler's  second  law,  if  A  be  the  area  described  by  the 
radius  vector,  dA/dt  is  constant, 

•••     f    or    Vtr\     (from  (5),  Art.  242) 


1  /   dy         dx\  „ 

=  7i\x-ji  —  2/~n=a  constant,  from 

2{   di         M)         (3),  Art.  242. 


FORCE  WHICH  ACTS  UPON   THE   PLANETS        479 
Differentiating  gives 


xC^- 

xdt2 

d2x 

yW2  = 

=  0; 

.     xY 

-yX  = 

:  0  (from 

(i)) 

7 

* 

X 
•     Y 

X 

■  -,     or 

y 

y  = 

X 

which  shows  that  the  axial  components  of  the  acceleration, 
due  to  the  attraction  acting  on  the  planet,  are  proportional 
to  the  coordinates  of  the  planet;  and  therefore  by  the 
parallelogram  of  forces,  the  resultant  of  X  and  Y  passes 
through  the  origin.  Hence,  the  forces  acting  on  the  planets 
all  pass  through  the  Sun's  center. 

(2)  From  the  first  of  Kepler's  laws  it  follows  that  the 
central  attraction  varies  inversely  as  the  square  of  the 
distance. 

The  polar  equation  of  the  ellipse,  referred  to  its  focus,  is 


a  (1  -  e2) 
9      1  +  e  cos  B 

1              1  -\-  e  cos  6 

or     -  =  u=  —Tz. ~-> 

p              a  (1  -  e2) 

which  by  differentiation  gives, 

d*u 
dd'2  "t" 

1 

U~  a(l-e2)' 

and, 
Art. 

therefore,  if  F  is  the  attraction  to  the  focus, 
242, 

by  (8), 

h2        1 

a  (1  -  e2)  p2 

Hence,  if  the  orbit  be  an  ellipse,  described  about  a  center  of  at- 
traction at  the  focus,  the  law  of  intensity  is  that  of  the  inverse 
square  of  the  distance. 

(3)    From  the  third  law  it  follows  that  the  attraction  of 
the  Sun  (supposed  fixed)  which  acts  on  a  unit  of  mass  of  each 


480  INTEGRAL  CALCULUS 

of  the  planets,  is  the  same  for  each  planet  at  the  same 
distance. 

By  Art.  242  (20),        T2  =  ^-a\ 

Since  by  the  third  law,  T2  varies  as  a3,  K  must  be  constant ; 
that  is,  the  strength  of  the  attraction  of  the  Sun  must  be  the 
same  for  all  the  planets.  Hence,  not  only  is  the  law  of  force 
the  same  for  all  the  planets,  but  the  absolute  force  is  the  same. 

The  third  law  shows  also  that  the  law  of  the  intensity  of 
the  force  is  that  of  the  inverse  square  of  the  distance. 

Since  the  planets  move  in  ellipses  slightly  different  from 
circles,  assume  for  simplicity  that  their  orbits  are  actually 
circles.  If  Ri,  R2,  Rs  are  the  radii  and  7\,  T2,  Ts,  the  respec- 
tive times  of  revolution  of  the  planets,  Kepler's  third  law 
may  be  written  as  follows: 

#i3      R23      #33  ,     , 

T?=  T72=  T?=   '  '  '   =  a  constant' 

The  expression  for  the  central  acceleration  of  motion  in 
a  circle  is  a  =  v2/R  =  4t2R/T2,  or  T2  =  4^/a  (Art.  70  (2)). 
Substituting  this  value  gives 
ai^Ri2  =  CI2R22  =  (I3R32  =  constant;     or    a  =  constant/7?2. 

Note.  —  Arts.  242,  243,  244,  are  based  on  the  discussion 
in  Bowser's  Analytic  Mechanics.  For  a  fuller  discussion,  see 
Tait  and  Steele's  Dynamics  of  a  Particle,  and  Percival  Frost's 
translation  of  Newton's  Principia,  Sec.  I,  II,  III. 

245.  Newton's  Verification.  —  The  greatest  of  Newton's 
achievements  is  considered  an  achievement  of  the  imagina- 
tion, his  conception  of  the  universality  of  natural  law.  At  an 
early  age  he  (in  1666)  conceived  with  his  far-reaching  mind 
the  then  daring  idea  that  the  sublime,  inscrutable,  central 
force  was  nothing  but  commonplace  gravity,  known  to  exist 
on  and  near  the  earth.  He  verified  his  idea  first  in  the  case 
of  the  moon.  He  discovered  that  the  same  acceleration  that 
controls  the  motion  of  an  object  near  the  earth  also  pre- 


NEWTON'S  VERIFICATION  481 

vented  the  moon  from  moving  away  in  a  rectilinear  path 
from  the  earth,  and  that  its  tangential  velocity  prevented 
it  from  falling  to  the  earth. 

Assuming  that  the  moon's  orbit  is  circular,  its  acceleration 
towards  the  earth  is  (by  Art.  70  (2)), 

a  =  v-  =  ^^  =  0.0089  ft. /sec.2, 

where  R  =  238,800  miles  and  T  =  27.32  days. 
From  the  law  of  inverse  squares : 

a  _  r2  _  r2  -  1 

~g~R2~  (60.267  r)2  ~  3632 ' 
a         32.089 


3632        3632 


0.0088  ft./sec2., 


where  32.089  is  the  value  of  g  on  the  earth  at  the  equator, 
and  R  is  60.267  times  r,  the  radius  of  the  earth. 

As  these  results  differ  by  only  To£ooth  of  a  foot,  the 
conclusion  is  that  the  centripetal  force  on  the  moon  in  its 
orbit  is  due  to  the  earth's  attraction,  acting  according  to 
the  law  of  inverse  squares. 

Owing  to  an  inaccurate  value  of  the  earth's  radius  which 
was  in  use  at  the  time  Newton  first  made  the  computation, 
the  result  then  obtained  seemed  to  show  that  the  law  of 
attraction  was  not  that  of  inverse  squares. 

Records  show  that  Newton,  although  unshaken  in  his 
belief,  laid  aside  his  calculations;  and  it  was  not  until 
thirteen  years  afterwards  that,  a  new  determination  of  the 
radius  having  been  made,  he  repeated  the  investigation  and 
found  the  verification  sought  for. 

Five  years  later  (in  1684)  he  was  induced  to  consider  the 
whole  subject  of  gravitation;  and  then  he  solved  the  supple- 
mentary problems  in  regard  to  the  attraction  of  a  sphere 
for  an  external  particle,  which  established  his  theory  — 
now  known  as  Newton's  Law  of  Universal  Gravitation. 


INDEX 


(Numbers  refer  to  pages) 


Acceleration,  19,  21 

angular,  92 

gravity,  90 

normal,  20,  89,  94 

resolution  of,  86 

tangential,  19,  86,  94 

total,  19 
Agnesi,  witch  of,  132 
Algebraic  functions,  9,  38 
Anti-derivatives,  263 
Anti-differential,  171 
Application  of  Taylor's  theorem, 

424 
Applications,  99,  355,  456 

beams,  229 

prismoid  formula,  284 
Approximate  formulas,  138,  249 

relative  rates  and  errors,  162 
Approximation  formulas,  421 
Arc,    differential  of,  17 

infinitesimal,  37 

length  of,  64 
Archimedes,  110,  142 
Area,  any  surface,  311 

as  an  integral,  223 

by  double  integration,  304,  308 

derivative  of,  205 

finite  or  infinite,  208 

hyperbola,  equilateral,  221 

positive  or  negative,  208 

under  a  curve,  206,  214 

under  derived  curves,  225 


Argument,  6 

Atmosphere,  limit  of,  378 
Attraction,  363 
Auxiliary  equation,  454 
theorems,  87 

Base,  6,  48,  56 

common,  54,  55 

Xaperian  or  natural,  49,  50,  54 
Beams,  229 
Binomial  theorem,  419 
Brachystochrone,  464 

Cable,  81 
Calculus,  1,  2,  17 

Differential,  3,  17,  100 

Integral,  3,  171 
Cardioid,  142,  217,  310,  311 
Catenary,  81,  140,  148,  245 

surface  generated  by,  293 

uniform  strength,  251 

volume  generated  by,  294 
Cauchy's  remainder,  416 
Center  of  curvature,  135,  142 

of  gravity,  332,  333,  334,  335 
Central  forces,  472 
Centrifugal  force,  90,  378 

railway,  466 
Centripetal  force,  90 
Centroid,  332 

Change  of  independent  variable, 
149 


483 


484 


INDEX 


Change,  rate  of,  1,  2,  5 

uniform  and  non-uniform,  14 
ChrystaFs  Algebra,  421 
Circle,    area   of,    217,    218,    261, 
309 

curvature  of,  135 

equation  of,  5 

length  of  circumference,  237 
Circular  functions,  64 

measure,  64,  70 

motion,  20,  88 

pendulum,  461 
Cissoid,  length  of,  237 

volume  generated  by,  295 
Coefficient,  differential,  24 

of  contraction,  469 

of  velocity,  468 
Comparison  test,  391 
Complementary  function,  453 
Complete  integral,  437 
Compound   interest    law,    9,    60, 

442 
Concavity,  122,  123 
Cone,  267,  270 
Conic  section,  475 
Conoid,  282 
Constant,  5 

absolute,  5,  6 

arbitrary,  5,  6 

of  integration,  173,  301 
Continuity,  7 
Continuous  functions,  7 

variables,  5,  7 
Convergence,  tests  for,  390,  393 
Convergent  series,  388 
Cosine,  differential  of,  64,  65 

graph  of,  119,  228 
Cubical  parabola,  140,  237 
Curvature,  133 

center  of,  135 

circle  of,  135 

radius  of,  135,  137,  140 


Curve,  8 

of  cord  with  load  uniform  hori- 
zontally, 241 

of  flexible  cord,  245 

polar,  123 
Curvilinear  motion,  460 
Curves,  derived,  118,  225 

integral,  227 

lengths  of,  236,  239 
Cusps,  114,  140,  146 
Cycloid,  105 

area  of,  217 

equation  of,  107 

evolute  of,  147 

length  of,  147,  237 

surface  generated  by,  293 

volume  generated  by,  294 
Cycloidal  pendulum,  464 
Cylinder,  267,  279,  280,  288,  290, 

291 
Cylindrical  coordinates,  324 

slice,  317 

surface,  318 

Damped  vibrations,  73,  441,  455 
Decreasing  function,  8 
Definite  integral,  206,  213 
Deflection  of  beams,  141 
Density  of  air,  62 

mean,  327,  374 
Dependent  variable,  6 
Derivative,  24,  25 

as  a  limit,  26 

as  slope  of  curve,  29 

of  area,  205 

partial,  153,  164 

successive,  84,  149 

total,  159 
Derived  curves,  118,  225 

function,  24 
Differential  Calculus,  3,  17 

coefficient,  24 


INDEX 


485 


Differential  equations,  436,  441 

of  sine  in  degrees,  69 
Differentials,  3,  15 

exact,  167 

inexact,  168 

partial,  152,  164 

successive,  84 

total,  156 
Differentiation,  2,  25 

of  series,  394 

rules  of,  64 
Discharge  from  an  orifice,  469 
Discontinuous  function,  7 

variables,  5 
Divergency  of  series,  391 
Double  integration,  304,  308,  324 

E,  modulus  of  elasticity,  230,  254 
e,  Napierian  base,  6,  50,  418 
Eccentricity,  475 
Elastic  curve,  230 
Elasticity,  modulus  of,  230,  254 
Elementary  principles,  178 
Elements,  261,  331 
Ellipse,  area  of,  218 

length  of  quadrant,  238,  398 
Ellipsoid,  292,  294 
Elliptic  integral,  400,  462,  463 
Empirical  equations,  10 
Energy  integral,  453 

kinetic,  92 
Equation  of  evolute,  144 

of  involute  of  circle,  146 

of  involute  of  cycloid,  146 

of  normal,  100 

of  tangent,  99 
Equations,  differential,  436 

exact  differential,  444 

homogeneous,  446 

linear,  448 

of  first  order,  449 

of  order  above  first,  449 


Equilateral  hyperbola,   140,   147, 
221 

Error  term,  421 

Errors,  percentage,  163 

Euler's  series,  397 

Evaluation  of   definite  integrals, 
213 
of  derivatives  of  implicit  func- 
tions, 435 
of  indeterminate  forms,  428 

Evolute,  143,  144,  256 

Evolution,  3 

Exact  differential  equations,  167, 
170,  444 

Examples,  illustrative,  31,  76,  160, 
174,  412 

Expansion  by  Maclaurin's  and  by 
Taylor's  Theorems,  411 
of  cosh  x/a  and  sinh  x/a,  248 
of  functions  in  series,  408 

Explicit  function,  6 

Exponential  function,"  9,  47 

Extended  law  of  the  mean,  40." 

Factor,  integrating,  445 
Falling  bodies,  21,  177,  297 
First  moment,  331 
Flexion,  21,  133 
Force,  central,  472 

centrifugal,  90,  378 

centripetal,  90 

concentrated,  355 

definition  of,  92 

distributed,  355 

variable,  472 
Forms,  indeterminate,  425,  428 

standard,  180, 181, 183,  203,  403 
Formula,  prismoid,  283 

projectile,  Helie's,  384 
Formulas,  38,  47,  64 

approximation,  421 

reduction,  194,  198 


486 


INDEX 


Frustum,  surface  and  volume  of 

any,  270 
Function,  6 

algebraic,  9,  38 

continuous,  7 

discontinuous,  7 

exponential,  9,  47 

hyperbolic,  9,  80 

inverse,  76,  82 

logarithmic,  9,  47 

of  a  function,  27 

power,  9,  10 

transcendental,  9 

trigonometric,  9,  64 

several  variables,  152 
Functional  relation,  7 
Fundamental  conception,  30 

condition  or  test,  112 

rule  for  applying  test,  115 

theorem,  261 

g,  acceleration  of  gravity,  90 
Gas,  formula  for,  160 
Gauge,  self-registering,  97 
Geometric    meaning    of    integral, 

204 
Geometric  progression,  9,  10 

series,  385 
Grade,  23 

Graphical  illustration,  113 
Graphs,  8,  80,  118 

cosine,  119,  228 

cycloid,  105 

sine,  71,  119,  228 

versine,  228 
Gravitation,  law  of,  363,  371,  481 

unit  of  mass,  376 
Gravity,  acceleration  of,  90,  91 

center  of,  332,  333,  334 
Gregory's  series,  397 
Guldin's  theorems,  336 
Gyration,  radius  of,  345 


Harmonic  law,  9 

motion,  94,  442,  457 

series,  389,  392 
Helie's  formula,  384 
Homogeneous  equations,  446 
Hyperbola,  equilateral,  140,  221 
Hyperbolic  functions,  9,  80 

logarithms,  216 
Hypocycloid,  140,  148 

Ideal  quantity,  2 
Ideas,  2 
Illustrations,  21 

typical,  118 
Illustrative  examples,  31,  76,  160, 

174,  412 
Implicit  function,  7 
Increasing  function,  8 
Increment,  12 
Indefinite  constant,  173,  301 

integral,  173,  301 
Independent  variable,  6,  149 
Indeterminate  forms,  425 
Inertia,  91 

moment  of,  230,  345,  346,  349, 
350,  352 

product  of,  348 
Inexact  differential,  168 
Infinite  series,  385 
Infinitesimal,  30,  36,  37 

arc  and  chord,  37 
Infinity,  36 
Inflexion,  point  of,  114,  120,  132, 

140 
Integral,  complete,  437,  453 

Calculus,  3,  171 

definite,  206 

energy,  453 

from  an  area,  219 

general,  227 

indefinite,  173,  301 

multiple,  297 


INDEX 


487 


Integral,  particular,  175,  453 
Integrals,  elliptic,  400,  462,  463 
Integrand,  171 
Integrating  factor,  445 
Integration,  2,  17,  170,  171 

double,  304,  308,  324 

constant  of,  173,  301 

parts,  194 

series,  394 

successive,  296,  303 

triple,  317,  321 
Intensity,  355 
Interchange  of  limits,  210 

of  order  of  differentiation,  165 

of  order  of  integration,  299,  301, 
306 
Interest,  compound,  law  of,  9,  60 
Inverse  functions,  9 

hyperbolic  functions,  81 

of  differentiation,  171 

trigonometric  functions,  76 
Involute,  143 

of  the  catenary,  248 
Involution,  3 

Jet,  liquid,  path  of,  467 

Kepler's  laws,  477 
Kinetic  energy,  92 

Lagrange's  remainder,  407 
Law,   compound  interest,    9,    60, 
442 

extended,  of  the  mean,  405 

of  organic  growth,  9,  62,  442 

of  the  mean,  401,  402,  403 

of  motion,  90,  359,  365 

parabolic,  9 

planetary  motion,  477 

universal  gravitation,  363,  371, 
481 
Lemniscate,  110,  142,  217,  326 


Lengths  of  curves,  236,  239 
Limit  of  a  sum,  259 

of  height  of  atmosphere,  378 

of  infinitesimal  arc,  37 
Limits,  25,  26,  30,  206,  208,  210 
Linear  equations,  448,  453 
Liquid  jet,  467 

pressure,  356 
Lituus,  124 
Locus,  8 
Logarithmic  differentiation,  56 

curve,  140 

functions,  47 

series,  395 

spiral,  142 
Logarithms,  54,  55 

common,  55 

hyperbolic,  216 

natural,  55 

Machin's  series,  397 
Maclaurin's  theorem,  407 

series,  408 
Mass,  91,  327 

unit  of,  376 
Maxima  and  minima,  111 

application  of  Taylor's  theorem, 
to,  424 

problems,  127 
Maximum  and  minimum,  111 
Mean  density,  327,  374 

law  of  the,  401,  402,  403 

value  of  a  function,  211 
Method  of  the  Calculus,  430 

of  limits,  30 
Modulus,  54 

of  elasticity,  230,  254 
Moment,  331,  345,  355 

of  inertia,  230,  350 

least  moment  of  inertia,  349 

of  inertia  for  parallel  axes,  346 

principal  moment  of  inertia,  348 


488 


INDEX 


Moments  of  area,  333,  352 

first,  331 

of  line,  334 

second,  344 

of  volume,  333 
Momentum,  92 
Motion,  circular,  88 

curvilinear,  460 

in  resisting  medium,  379,  383, 
459 

planetary,  477 

rectilinear,  456 

second  law  of,  90 

simple  harmonic,  94,  442,  457 

third  law  of,  90,  359,  365 

Napierian  or  natural  base,  49,  50, 
54 

logarithms,  55,  70,  216 
Newton,    4,    90,    359,    363,    371, 

480 
Normal  acceleration,  20,  89,  94 
Normal,  99 

polar,  109 
Notation  for  functions,  6 
Number  e,  6,  50,  418 

7T,  6,  50,  398,  418 

Orbit,  472,  475 

Order  above  the  first,  449 

first,  449 

of  a  differential  equation,  436 
Ordinary     differential     equation, 

436,  444 
Organic  growth,  9,  62,  442 
Orifice,  discharge  from,  469 
Oscillating  series,  385,  397 

7T,  6,  50,  398,  418 
Pappus,  theorem  of,  336 
Parabolic  cable  or  cord,  241 
law,  9 


Partial  derivations,  153,  164 

differentials,  152,  164 
Particular  integral,  175,  453 

values,  6 
Parts,  integration  by,  194 
Path  of  a  projectile,  381 

of  a  liquid  jet,  467 
Pendulum,  simple  circular,  461 

cycloidal,  464 
Percentage  rate,  56 

error,  163 
Period,  96 
Periodic  time,  477 
Plane,  tangent,  154 
Plane  areas,  304,  308 
Planetary  motion,  477,  478 
Points  of  inflexion,  114,  120,  132, 

140 
Polar  curve,  8,  123 

moment  of  inertia,  345 

subtangent,  subnormal,  108 
Power  form,  183 

formula,  44 

function,  9,  10 

series,  386,  393 
Pressure  of  air,  62 

liquid,  356 
Primitive  of  differential  equation, 

437 
Principles,  371,  480 
Prismoid  formula,  283,  284 
Probability  integral,  418 
Problems,   maxima   and   minima, 

127 
Process,  summation,  261 
Product  of  inertia,  348 
Progression,  9,  10 
Projectile,  path  of,  381 

Quadrature  of  curves,  443 
Quotient,  differential  of,  39,  42 
limit  of,  26,  35 


INDEX 


489 


Radian,  64,  70 
Radium,  63 

Radius   of   curvature,    135,    137, 
140 

of  gyration,  345 
Railway,  centrifugal,  466 
Range  formula,  Helie's,  384 
Rate,  18,  19,  21 

of  change,  1,  2 

percentage,  56 

relative,  56,  162 
Rectification  of  curves,  237 
Rectilinear  motion,  456 
Relative  error,  59,  162 

rate,  56,  162 
Remainder,  Cauchy's,  416 

Lagrange's,  407 
Remarks,  27,  70 
Replacement  theorem,  37 
Representation  of  functional  re- 
lation, 7 

volume  by  area,  269 
Resisting  medium,  379,  383,  459 
Rotation,  92 

Secant,  29,  65,  68 
Second  derivative,  84,  86 

moments,  344 
Semi-cubical  parabola,  145,  237 
Separable  variable,  446 
Separated    differential    equation, 

444 
Separation  into  parts,  210 
Series,  infinite,  385 

absolutely  convergent,  388 

convergent,  385,  393 

divergent,  385 

Euler's,  397 

for  e,  50 

Gregory's,  397 

geometric,  385 

harmonic,  389,  392 


Series,    integration   and   differen- 
tiation of,  394 

logarithmic,  395 

Machin's,  397 

Maclaurin's,  408 

Newton's,  397 

oscillating,  385,  397 

power,  386,  393 

Taylor's,  409 
Shear,  229 
Shooting  point,  114 
Significance  of  area,  223 
Simple  circular  pendulum,  461 

harmonic  motion,  94,  442,  457 
Sine  curve,  72 

differential  of,  64,  65 

graph  of,  71,  119,  228 

ratio  to  arc,  66,  67 
Slope,  18,  23,  99 

rate  of  change  of,  21 
Solids  of  revolution,  288 

by  double  integration,  321 
Solution  of  a  differential  equation, 
436 

of  s  =  a  sinh  x/a,  250 
Speed,  1,  18,  19 
Sphere,  268,  274,  285,  313,  323 
Spherical  shell,  368,  371 
Spheroid,  294 
Spiral  of  Archimedes,  110,  142 

logarithmic,  110,  142 
Standard  forms,  180,  182,  203 
Statical  moment,  345,  359 
Stiffness  of  beams,  128 
Strength  of  beams,  127 
Subnormal,  100,  108 
Subtangent,  100,  108 
Successive  derivatives,  84,  149 

differentials,  89 

differentiation,  84 

integration,  296,  300 
Summation  process,  261 


490 


INDEX 


Summation  process,  approximate 

and  exact,  262 
Surface  of  any  frustum,  270 

of  revolution,  288 
Suspension  bridge,  244 

Tangent,  equation  of,  99 

plane,  154 

polar,  109 
Tangential  acceleration,  19,  86,  94 
Taylor's  theorem,  405,  406,  407 

series,  409 
Test,  comparison,  391 

for  convergence,  390 

ratio,  391 
Theorem,  replacement,  37 

binomial,  419 

finite  differences,  402 

Maclaurin's,  407 

Taylor's,  405,  406,  407 
Theorems,  auxiliary,  124 

of  limits,  26 

of  mean  value,  401,  402 
Theorems  of  Pappus  and  Guldin, 

'336 
Tide  gauge,  97 
Time,  periodic,  477 

rate  of  change,  19 
Total  derivative,  159 

differential,  156 
Tractrix,  254 
Transcendental  functions,  9 

numbers,  6,  50 
Trapezoid,  343 

Trigonometric  functions,  9,  64 
Triple  integration,  317,  32K 
Typical  illustrations,  118 


Uniform  change,  14 

speed  or  velocity,  20 
Unit  of  force,  91 

of  mass,  91,  376 
Universal   gravitation,    363,   371, 

481 
Use  of  standard  formulas,  182 

Value,  mean,  211 
Variable,  5 

continuous,  5 

dependent,  6 

discontinuous,  5 

independent,  6,  149 

separable,  446 
Vector  quantity,  21,  355 
Velocity,  18 

angular,  92 

average,  22,  23 

constant,  20 

instantaneous,  1 

on  a  curve,  94 

tangential,  94 

uniform,  20 
Verification,  Newton's,  480 
Vibrations,  damped,  73 
Volume  by  an  area,  268 

of  frustum,  270 
Volumes,  267 

by  double  integration,  324 

by  triple  integration,  317,  321 

Water  pressure,  357 
Wave  curve,  72 
Wetted  perimeter,  129 
Witch  of  Agnesi,  132 
Work,  167 


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